# Completing the Square

Lesson Features »

Lesson Priority: High

Algebra One $\longrightarrow$
Intro to Quadratics $\longrightarrow$

Objectives
• Learn how to solve a quadratic equation by completing the square
• Practice the simple case of this technique with quadratics that have a leading coefficient of 1
• Practice the general case of this technique with quadratics that have leading coefficients other than 1
• Know how to handle all steps of the process in cases when fraction coefficients are present in the initial problem
Lesson Description

This technique for solving quadratics is often puzzling to students, since it involves some strange equation manipulation. But this technique has a few important uses in the future, and particularly is a topic that gets easier with practice!

Practice Problems

Practice problems and worksheet coming soon!

## A Massive Manipulation Game

Completing the Square is a technique that many students are confused about at first, for two reasons. First, the actual technique looks like 100% smoke and mirrors tricks, since it entirely involves moving terms around in a clever way. Second, even after you see the steps and know how to proceed, it is not at all obvious why this manipulation technique is useful in any way. Read on to get closure on both of these issues, as well as see what kind of exam questions your teacher will throw at you.The main idea of completing the square is that we can take a quadratic equation and solve it using neither factoring methods nor the quadratic formula. The way we will proceed is to make one side of the equation a perfect square trinomial, and then take the square root of each side and solve for $x$.

## Remembering Perfect Square Trinomials

First, let's recall what a perfect square trinomial even looks like. $x^2 - 6x + 9$ is a perfect square trinomial because$$(x-3)^2=x^2 -6x + 9$$Likewise, $x^2 + 5x - 7$ is not a perfect square trinomial, because there is no way to write this trinomial in the factored form $(x + a)^2$.In short, from what we learned in the past lesson on Perfect Square Trinomials », we know that we're dealing with a perfect square trinomial either if we start off with factored form that looks like $(x-a)^2$, or if we start off with standard form that we can factor into $(x-a)^2$ form. There is another property that standard form perfect square trinomials have, and it's key to the Completing the Square method.
Theorem: Perfect Square TrinomialsA perfect square trinomial with quadratic coefficient of $1$ has the following property: the square of one-half of the linear term coefficient is equal to the constant term.
This may sound complicated in words, but it is much more manageable in practice. For example, in the perfect square trinomial we just looked at, $x^2 -6x + 9$, half of $(-6)$ is $(-3)$, and $(-3)^2$ is $9$:$$x^2 -6x + 9 \longrightarrow \left(\frac{1}{2} \, (-6)\right)^2 = 9$$Here are a few other examples:$$(x+5)^2 = x^2 + 10x + 25 \longrightarrow \left(\frac{1}{2} \, (10)\right)^2 = 25$$$$(x-2)^2 = x^2 - 4x + 4 \longrightarrow \left(\frac{1}{2} \, (-4)\right)^2 = 4$$$$(x+15)^2 = x^2 + 30x + 225 \longrightarrow \left(\frac{1}{2} \, (30)\right)^2 = 225$$This property is simply a consequence of what perfect square trinomials are. But it's your job to remember this, because this fact is of paramount importance to executing the Completing the Square process.
Remember!
You can identify a perfect square trinomial by checking that squaring one-half of its middle (linear) term gives you the last (constant) term. It sounds unnecessarily complicated relative to factoring, but it is the best approach when you need to create a perfect square. Also, while factoring is more comfortable and a great way to check your work, it doesn't hold up for speed and accuracy when we eventually have to work with fraction coefficients.
Now that we remember what perfect square trinomials are, and know how to identify standard form perfect square trinomials using the $1/2$ middle squared trick, we can proceed with the major, repeatable steps for any problem that instructs us to use the Completing the Square method.
You Should Know
There are two major cases where the method of Completing the Square is required:Re-writing a Standalone ExpressionThis is where we'll take a quadratic expression such as$$ax^x + bx + c$$and both add and subtract equal amounts to re-write the quadratic as a perfect square binomial plus or minus a constant. This skill is useful in Algebra Two when we will need to express quadratic equations in various forms. One form, called vertex form, requires use of the Complete the Square method in order to turn any other quadratic form into vertex form. We won't go that far in this lesson but we will learn how to complete the square of a standalone expression.Solving Quadratic EquationsWhen you're asked to solve a one-variable quadratic equation using the Completing the Square method, you will typically be starting out with an equation of the form$$ax^2 + bx + c = d$$where $a$, $b$, $c$, and $d$ are number coefficients. Beginning and intermediate problems typically have integer coefficients, while advanced problems may have fraction coefficients. Also, the process is much, much simpler when $a$ is $1$. We will learn this process for situations where $a$ is $1$, and practice it a few times, before moving on to the case when $a$ is not $1$.Regardless of which type of question we are answering, there is a noticeable difficulty difference between problems that do and do not have a $a$ coefficient other than $1$. We'll practice both in this lesson.But let's not run before we can crawl - first we will learn how the Complete the Square process works.

## Complete the Square - Expressions

The idea behind completing the square is that whatever $ax^2$ and $bx$ terms you have, you can figure out what the correct $c$ should be to make a perfect square. Add whatever constant to the expression that you need to achieve the correct $c$ value, and then subtract the same amount so that you effectively added $0$.Let's see an example to understand how it works.Example 1Complete the square of the following expression.$$x^2 + 20x +56$$$\blacktriangleright$ Because of the one-half squared rule we observed above for perfect square trinomials, we know that $x^2 + 20x$ can only be part of a perfect square if the constant term is one-half of $20$, then squared (which is $10^2$ or $100$). However, we don't have $100$ - we have $56$. But if we add $44$ and subtract $44$ simultaneously, we can group the pieces such that we have a perfect square trinomial.$$x^2 + 20x +56 + 44 - 44$$$$\longrightarrow \left(x^2 + 20x +56 + 44\right) - 44$$$$\longrightarrow (x + 10)^2 - 44$$There you have it - we have completed the square. By adding and subtracting the right number, we can rewrite the expression as a perfect square and an added or subtracted constant. It's all possible to do because of our ability to goal-seek the correct constant that we need using the "one-half $b$ then squared" rule of perfect square trinomials.

Next we'll apply this skill to solving equations, and then move on to more difficult cases such as when $b$ is an odd number, or when $a$ is not $1$.

## Complete the Square to Solve Equations

Now let's use the idea of creating perfect square trinomials to solve equations.Step 1 - Create a Perfect Square TrinomialThe quadratic we will be given in a problem will not be a perfect square - but we can make it a perfect square by adding and subtracting the right number on one side. Here is the first example we will work:$$x^2 -6x -2 = 5$$(1)Look at the left side. We want that to be a perfect square trinomial. We also know that if we have$$x^2 - 6x + \boxed{\, ? \,}$$That the missing number that would make it a perfect square integer is $9$, via the one-half middle squared rule. The typical way we will proceed is to simultaneously add and subtract that magic number on the left side.$$x^2 -6x - 2 + 9 - 9 = 5$$$$\longrightarrow \left( x^2 -6x + 9 \right) - 9 - 2 = 5$$(2)By adding and subtracting $9$ at the same time, we are in effect adding $0$ to the left side. Since we are adding $0$, we have not changed the equation. But now, we can group the perfect square trinomial as a separate item.
Pro Tip:
Note that some students will prefer to get from result (1) to result (2) with two steps: adding the left-side constant to both sides, then adding the appropriate additional constant to each side to create the perfect square. If we took this approach, we would have gone from $x^2 - 6x - 2 = 5$ to $x^2 -6x = 7$, and THEN added the $9$ to each side to get $x^2 -6x + 9 = 16$. The benefit is that it reduces the likelihood of making an arithmetic error. The detriment is that it's one more step to write. My method is preferred however because then, every time you Complete the Square, whether it is for solving equations which we will learn in this lesson, or for Vertex Form conversion which we will see in the next lesson, you will be doing the process in a consistent way.
Step 2 - Factor the Perfect SquareNow that we have a perfect squared trinomial on the left, we can factor it instantly without having to work to figure out its factored form. This is because a perfect square trinomial will always factor in the form $\left(x + \frac{b}{2}\right)^2$. For this problem, our $b$ value was $(-6)$, so $b/2$ is $(-3)$, and without having to work it out long-form, we can say that $x^2 - 6x + 9$ factors as $(x-3)^2$. This leaves us with$$\left( x^2 -6x + 9 \right) - 9 - 2 = 5$$$$\longrightarrow (x-3)^2 - 9 - 2 = 5$$(4)If you forget the one-half middle trick for fast factoring, you can of course find the factored form fairly quickly via traditional quadratic factoring techniques. However, you absolutely must make sure you write the factored form as a square, otherwise you're apt to forget how to proceed from here. E.g. if you obtain $(x-3)(x-3)$ from factoring $x^2 -6x + 9$, you need to write it as $(x-3)^2$ to proceed.Step 3 - Isolate the Perfect SquareThis step is very important. In order to proceed correctly you need to make sure the $(x-a)^2$ perfect square is isolated. Sometimes that will take very few steps, such as in this problem, and sometimes that will take a few more. Here, we need only move the $-9$ and $-2$ constants to the right side. We'll combine like terms while we're at it.$$\left(x-3\right)^2 - 9 - 2 = 5$$$$\longrightarrow \left(x-3\right)^2 = 16$$(5)Step 4 - Square Root Both SidesTake the square root of both sides of (5). Since the left side is a squared object, taking the square root of it implies that the result may have been positive or negative. In other words, just like $\sqrt{x^2} = \pm x$, our equation will become$$\sqrt{\left(x-3\right)^2} = 16 \,\,\longrightarrow \,\, x-3 = \pm 4$$(6)If you need a refresher on why this $\pm$ symbol is necessary in this setting, revisit the lesson on Solving Quadratics with Square Roots ».
Remember!
If you square root a squared object in an equation setting, you absolutely must include both the positive and negative possible results (most succinctly notated via $\pm$). If you only include the $+$ result, you will only get one answer, even though any quadratic you solve should yield two answers.
Step 5 - Isolate $x$At this point, you will always have the same one small step to polish off the problem: Isolate $x$ by moving the constant on the left side to the right side, and then clean up the result, including simplifying the square root if possible. In this case:$$x-3 = \pm 4$$$$+3 \,\,\,\,\, = \,\,\,\,\, +3$$$$x = 3 \pm 4 \Longrightarrow x = -1, \,\, 7$$The solution to our problem is $x = -1$ or $x = 7$.Let's do another example, which we'll handle with the same five steps. Try it yourself first!

Example 2$$x^2 + 8x + 1 = 34$$
Show solution
$\blacktriangleright$ Examining the left side quadratic reveals that the required constant to make a perfect square is $16$. This is a result of the $(b/2)^2$ trick, since $(8/2)^2 = 16$. Therefore we will add and subtract $16$ on the left side (Step 1).$$x^2 + 8x + 1 + 16 - 16 = 34$$$$\longrightarrow \left(x^2 + 8x + 16\right) + 1 - 16 = 34$$We have now created a perfect square trinomial on the left side. Next, let's factor that trinomial (Step 2), recalling that $b/2$ is the number that will end up in the factored form:$$\left(x-4\right)^2 + 1 - 16 = 34$$Isolate the squared expression (Step 3) before proceeding.$$\left(x-4\right)^2 = 49$$Next up, we'll take the square root of both sides of the equation (Step 4).$$\sqrt{(x-4)^2} = \sqrt{49} \longrightarrow x-4 = \pm 7$$Finally, we'll finish off the problem by isolating $x$ and computing the two solutions (Step 5).$$x - 4 = \pm 7 \longrightarrow x = 4 \pm 7$$$$x = 4 \pm 7 \Longrightarrow x = -3, \,\, 11$$Therefore, the solution to the equation is $x=-3$ or $x=11$.
Things work out the same but feel more complicated when the linear term is an odd number, because fractions get involved. Try out this next example following the same four steps, but don't be intimidated when fractions appear - it all works exactly the same way!

Example 3$$x^2 - 7x - 5 = 25$$
Show solution
Step 1$\blacktriangleright$ We want a perfect square on the left side. To create one, use the one-half middle trick: we need to add and subtract $\left(\frac{1}{2} \cdot 7 \right)^2$ or $\frac{49}{4}$ to the left side:$$x^2 - 7x -5 + \frac{49}{4} - \frac{49}{4} = 25$$$$\Rightarrow \left(x^2 - 7x + \frac{49}{4} \right) -\frac{49}{4} - 5 = 25$$Step 2Now the grouped trinomial on the left side is guaranteed to be a perfect square trinomial, and furthermore, we know it factors as $(x-a)^2$ where a is the one-half middle number, or in this case, $7/2$. Rewriting the equation with factored form yields$$\left(x-\frac{7}{2}\right)^2 -\frac{49}{4} - 5 = 25$$Step 3Isolate the squared expression by moving the extra constants on the left side to the right side. To make the quickest work of this task, convert everything to have a denominator of $4$.$$\left(x-\frac{7}{2}\right)^2 -\frac{49}{4} - \frac{20}{4} = \frac{100}{4}$$$$\left(x-\frac{7}{2}\right)^2 = \frac{169}{4}$$Step 4Square root both sides of this equation, remember that the square root of a fraction is done by taking the square root of each the numerator and the denominator.$$\sqrt{\left(x-\frac{7}{2}\right)^2} = \sqrt{\frac{169}{4}}$$$$\left(x-\frac{7}{2}\right) = \pm \frac{13}{2}$$Step 5Finally, isolate $x$ and obtain the answers.$$x = \frac{7}{2} \pm \frac{13}{2}$$$$x = 10, \,\, -3$$

## Mr. Math Makes It Mean

Most teachers will give you at least a few problems on a quiz or test where there is a constant other than $1$ on the quadratic term. We will be able to follow the same process, but there is just a little more work to do.Here's our example to learn by:Example 4$$3x^2 + 4x + 2 = \frac{102}{3}$$Step 1In order to create a perfect square trinomial using our $(b/2)^2$ trick, we need to work with a quadratic that has an $a$ coefficient of $1$. Since we do not have one, we have to factor out the leading coefficient so that we will have such an object to work with, even though the leading coefficient doesn't factor "nicely" out of the linear term.$$3x^2 + 4x + 2 = \frac{102}{3}$$$$\longrightarrow 3 \left(x^2 + \frac{4x}{3}\right) + 2 = \frac{102}{3}$$Now, when we create a perfect square trinomial, we will do so inside the parenthesis. Examining the quadratic inside the parenthesis (which is currently $x^2 + 4x/3$) we see that the $b$ term is $4/3$. Half of $4/3$ is $2/3$, and $2/3$ squared is $4/9$. We will add and subtract this amount, again, working inside the parenthesis.$$3 \left(x^2 + \frac{4x}{3} + \frac{4}{9} -\frac{4}{9} \right) + 2 = \frac{102}{3}$$Step 2Now we'll factor the perfect square trinomial we created. We are still keeping it inside the parenthesis for now.$$3 \left(\left(x+ \frac{2}{3}\right)^2 - \frac{4}{9} \right) + 2 = \frac{102}{3}$$Step 3We must isolate the $(x + 2/3)^2$ term. This is the part you need to pay attention to, because students rush at this point, when instead they need to be careful. The term $-\frac{4}{9}$ in the parenthesis comes out of the parenthesis as $-\frac{4}{3}\,\,$, because it needs to be multiplied by the front coefficient of $3$. A quick look at this symbolically to help say this a different way:$$3(a + b + c + d) = 3(a + b + c) + 3d$$Now our equation looks like$$3 \left(x+ \frac{2}{3}\right)^2 - \frac{4}{3} + 2 = \frac{102}{3}$$And, cleaning up the constants$$3 \left(x+ \frac{2}{3}\right)^2 + \frac{2}{3} = \frac{102}{3}$$$$\rightarrow 3 \left(x+ \frac{2}{3}\right)^2 = \frac{100}{3}$$$$\rightarrow \frac{\cancel{3} \left(x+ \frac{2}{3}\right)^2}{\cancel{3}} = \frac{\frac{100}{3}}{3}$$$$\rightarrow \left(x+ \frac{2}{3}\right)^2 = \frac{100}{9}$$Step 4Now let's take the square root of both sides.$$\sqrt{\left(x + \frac{2}{3}\right)^2} = \sqrt{\frac{100}{9}}$$$$x + \frac{2}{3} = \pm \frac{10}{3}$$Step 5Move the $2/3$ over to the left side to isolate $x$ and finish the problem.$$x = -\frac{2}{3} \pm \frac{10}{3}$$$$\boxed{\Rightarrow x = -4, \,\, \frac{8}{3}}$$

Remember!
This example had a nice square root result for the denominator on the right side, but if it doesn't, you need to rationalize. You should do it right away. If you don't rationalize you will very likely lose points.
Although in practice you should expect to see a healthy mix of difficulty in practice problems and exams, this process is rigid - the same 5 steps always apply. It's a matter of how much algebra clean-up is required.

## Put It To The Test

Because this process is repetitive and structured, a little practice will set you straight really quickly for a quiz or test. Make sure you can do the problems that require square root simplification and / or working with fractions, as teachers tend to universally give a few on tests to sift out the top scoring students.Examples 5 to 7: rewrite each expression as the sum of a squared binomial and a constant, using the Completing the Square method.

Example 5$$x^2 - 12x - 12$$
Show solution
$\blacktriangleright$ Focusing on the terms with variables, the one-half squared rule tells is that we need the constant to be $36$. Let's add and subtract $36$ to the expression, so that we don't change its value.$$x^2 - 12x +36 - 12 - 36$$$$\longrightarrow \left[x^2 - 12x + 36\right] - 48$$Now that we've created a perfect square trinomial, factor it to achieve the task.$$\longrightarrow (x - 6)^2 - 48$$

Example 6$$x^2 - 9x + 1$$
Show solution
$\blacktriangleright$ Just like the last problem - focus on the linear term to use the one-half squared rule.$$x^2 - 9x + \frac{81}{4} + 1 - \frac{81}{4}$$$$\left(x^2 - 9x + \frac{81}{4} \right) -\frac{77}{4}$$$$\left(x - \frac{9}{2} \right) - \frac{77}{4}$$

Example 7$$2x^2 - 5x + 12$$
Show solution
$\blacktriangleright$ We need to be a little more careful when there is a coefficient other than $1$ on the quadratic term. Factor it out from both variable terms, regardless of the linear term's coefficient. Leave the constant out of the parenthesis.$$2\left(x^2 - \frac{5x}{2} \right) + 12$$Now, complete the square inside the parenthesis.$$\longrightarrow 2\left(x^2 - \frac{5x}{2} + \frac{25}{16} - \frac{25}{16} \right) + 12$$Distribute the $2$ in front to only the $-25/16$:$$2\left(x^2 - \frac{5x}{2} + \frac{25}{16} \right) - \frac{25}{8} + 12$$$$2\left(x^2 - \frac{5x}{2} + \frac{25}{16} \right) + \frac{71}{8}$$Finally, factor the perfect square trinomial.$$2 \left( x^2 - \frac{5}{4} \right) + \frac{71}{8}$$
You Should Know
Fraction coefficients come up more frequently for standalone expressions than for equation solving. This is because in equations, we can often multiply both sides of the equation by a constant to "clear" the fractions. This fraction business isn't as common for quizzes as the integer cases, but it still can appear for hard or bonus questions.
Examples 8 through 13: solve each equation using the Completing the Square method.

Example 8$$x^2 - 10x + 4 = 27$$
Show solution
$\blacktriangleright$ The magic number to add and subtract for this one is $25$. We'll add and subtract it on the left side, and then factor the resulting perfect square trinomial.$$x^2 - 10x +25 - 25 + 4 = 27$$$$\left(x - 5\right)^2 -21 = 27$$Move the remaining $21$ to the right to isolate the squared expression, and then take the square root of both sides:$$\left( x - 5 \right)^2 = 48$$$$\sqrt{\left( x - 5 \right)^2} = \sqrt{48}$$$$x - 5 = \pm \sqrt{48}$$Finally, we'll isolate $x$ and write the solution. Don't forget to simplify $\sqrt{48}$, as most teachers will dock points if you don't!$$x - 5 = \pm 4\sqrt{3}$$$$x = 5 \pm 4\sqrt{3}$$Our two answers are $5 + 4\sqrt{3}$ and $5 - 4\sqrt{3}$.

Example 9$$x^2 + x - 8 = 10$$
Show solution
$\blacktriangleright$ The $b$ coefficient is $1$, so $(b/2)^2$ is $1/4$.$$x^2 + x + \frac{1}{4} - \frac{1}{4} - 8 = 10$$Factor the perfect trinomial:$$\left(x^2 + x + \frac{1}{4} \right) - \frac{1}{4} - 8 = 10$$$$\left(x-\frac{1}{2}\right)^2 - \frac{1}{4} - 8 = 10$$Next we need to isolate the perfect square$$\left(x-\frac{1}{2}\right)^2 = 10 + 8 + \frac{1}{4} = \frac{73}{4}$$Take the root of both sides, then isolate $x$ for the final answers$$\sqrt{\left(x - \frac{1}{2}\right)^2} = \sqrt{\frac{73}{4}}$$$$x - \frac{1}{2} = \frac{\sqrt{73}}{2}$$$$\boxed{x = \frac{1 \pm \sqrt{73}}{2}}$$

Example 10$$x^2 + 16x + 100 = 309$$
Show solution
$\blacktriangleright$ Here are the steps, presented without description. The same 5 step approach applies that we've been doing the whole time.$$x^2 + 16x + 64 - 64 + 100 = 261$$$$\left(x+8\right)^2 - 64 + 100 = 261$$$$\left(x+8\right)^2 = 225$$$$\sqrt{\left(x+8\right)^2} = \sqrt{225}$$$$x + 8 = \pm 15$$$$x = -8 \pm 15$$$$\boxed{x = -23, \,\, 7}$$

Example 11$$-4x^2+12x - 2 = -17$$
Show solution
$\blacktriangleright$ Remember the two little nuances involved with a Complete the Square problem that has a leading coefficient other than $1$: you must factor out the coefficient, and you must work inside the parenthesis when creating your perfect square trinomial.$$-4 \left(x^2 - 3x \right) - 2 = -17$$The "$b$" we will work with to create the perfect square is $3$, so the number to add and subtract is $9/4$.$$-4 \left(x^2 - 3x +\frac{9}{4} - \frac{9}{4} \right) - 2 = -17$$$$-4 \left(\left( x-\frac{3}{2} \right)^2 - \frac{9}{4} \right) - 2 = -17$$Now we will work toward isolating the squared term, taking the square root of both sides, and finishing the problem. Note that the $-9/4$ will come out of the parenthesis as $+9$, due to the $-4$ front coefficient.$$-4 \left( x-\frac{3}{2} \right)^2 + 9 - 2 = -17$$$$-4 \left( x-\frac{3}{2} \right)^2 = -24$$$$\left( x-\frac{3}{2} \right)^2 = 6$$$$\sqrt{\left( x-\frac{3}{2} \right)^2} = \sqrt{6}$$$$x-\frac{3}{2} = \pm \sqrt{6}$$$$x = \frac{3}{2} \pm \sqrt{6}$$$$\boxed{x = \frac{3 \pm 2\sqrt{6}}{2}}$$

Example 12$$x^2 + 12x - 4 = -60$$
Show solution
$\blacktriangleright$ The process is once again the same. See if you can line up the work below with the process steps.$$x^2 + 12x + 36 -36 - 4 = -60$$$$\left( x + 6 \right)^2 -36 -4 = -60$$$$\left( x + 6 \right)^2 = -20$$$$\sqrt{\left( x + 6 \right)^2} = \sqrt{-20}$$$$x + 6 = \pm 2i \sqrt{5}$$$$x = -6 \pm 2i \sqrt{5}$$One small difference in this problem from others we've seen in this lesson is that this solution involved the imaginary number. This isn't a substantial difference in terms of the work you need to do, but rather something to be aware of.
You Should Know
Just like the Quadratic Formula can lead to imaginary answers, so can any other method of solving quadratic equations. The frequency in which you can expect to see imaginary answers varies greatly from teacher to teacher.

Example 13$$\frac{5x^2}{3} + \frac{3x}{8} + \frac{17}{6} = \frac{21}{2}$$
Show solution
$\blacktriangleright$ Here we may have a better time starting out if we try and eliminate some of the fractions. We could proceed as normal but the technique of clearing fractions, while optional, is sometimes helpful. A multiplier of $24$ will clear all fractions.$$24 \cdot \left[ \frac{5x^2}{3} + \frac{3x}{8} + \frac{17}{6} \right] = 24 \cdot \left[ \frac{21}{2} \right]$$$$40x^2 + 9x + 68 = 252$$Now we'll need to factor out the leading coefficient, as we've seen done previously. There is no other way to proceed, and more fractions will be created. So remember that we cleared fractions to make it a little more digestible, not to get rid of them from the problem completely. There would have been many more fractions if we didn't.$$40 \left( x^2 + \frac{9x}{40}\right) + 68 = 252$$The magic number here is $81/6400$.$$40 \left(x^2 + \frac{9x}{40} + \frac{81}{6400} - \frac{81}{6400}\right) + 68 = 252$$$$40 \left(\left(x + \frac{9}{80}\right)^2 - \frac{81}{6400}\right) + 68 = 252$$Now we need to isolate the squared expression:$$40 \left(x + \frac{9}{80}\right)^2 - 40 \cdot \frac{81}{6400} + 68 = 252$$$$40 \left(x + \frac{9}{80}\right)^2 - \frac{81}{160} = 184$$$$40 \left(x + \frac{9}{80}\right)^2 = \frac{29521}{160}$$$$\left(x + \frac{9}{80}\right)^2 = \frac{29521}{6400}$$Now we take the square root of each side$$\sqrt{\left(x + \frac{9}{80}\right)^2} = \sqrt{\frac{29521}{6400}}$$$$x + \frac{9}{80} = \frac{\sqrt{29521}}{80}$$and isolate $x$ to finish the problem$$x = \frac{9 \pm \sqrt{29521}}{80}$$It is worth mentioning that $29521$ does not have a simplified root because it is nearly prime (its only factors are $53$ and $557$). But even if it did break down, you will probably not be expected to do it for such large results. That said, you usually won't get such large results. Of course, this problem was purposefully complicated in terms of coefficients so that you can see how bad it can get on quizzes. Most problems you'll do on this topic will be a bit more manageable.
You Should Know
Completing the square is not the only way to solve quadratic equations, and the more complicated it looks, the more likely that the Quadratic Formula » is the most efficient way to solve the equation. However, at the demands of the teacher, we often have to do one or two of these complicated ones using the Complete the Square method. At the end of the day, when working on a graded assignment, you must follow the instructions that are given to you. If a problem tells you to solve an equation using Completing the Square, you must.

Lesson Takeaways
• Understand what kind of equations and situations we use the Complete the Square method for
• Recall what perfect square trinomials are and be able to manipulate a quadratic into becoming one
• Be able to execute the Complete the Square method for basic cases when $a$ is 1
• Master the method for more involved cases, such as fraction coefficients and the $a$ coefficient being different than $1$
• Turn in answers that are simplified sufficiently, including a single fraction answer when applicable, or simplified square roots when applicable
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Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

Key Lesson Sections

Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.