3x^2+15x=108 I've been on this problem for days now for my summer work and I've googled how to solve it and I still don't understand help would be very appreciated

Completing the Square. Standard form of a quadratic is ax

^{2}+bx+c=0.The first step in CtS is to subtract the "c"... but this has already been done for you.

3x

^{2}+15x=108Next, divide all of the terms by "a"... here
a=3.

3x

^{2}/3+15x/3=108/3x

^{2}+5x=36Next, we actually "complete the square." We take the expression on the left and turn it into a perfect square trinomial by finding a new constant (number with no variable). To keep the equation balanced, we will add this value to both sides of the equation.

Now, how to find it? Take your new "b" value (in this case:
+5... the sign will be important when we factor...), divide it by 2, then square it. Word to the wise: keep it "messy" on the left.

x

^{2}+5x+(+5/2)^{2}=36+6.25 Note: it's your choice to use 6.25 or leave it as +(5/2)^{2}on the right.Now, factor it. If you kept it messy on the left, take take a variable and what's in the parentheses to create your binomial. You can check this with actual factoring techniques, and you'll see that it works.

(x+ 5/2)

^{2}=42.25Use the Square Root Property. Don't forget the ±roots when you take the square root of both sides of the equation.

x+ 5/2 = ±√42.25

x+ 5/2 = ±6.5

Solve for your variable. Remember, you basically have two equations here (because of the ±). If you need to see them, they are:

x+ 5/2 = 6.5 x+ 5/2 = -6.5

- 5/2 -5/2 -5/2 -5/2

x = 4 x = -9

Regardless of which method you use to solve this particular equation, we should all get x=4 and x=-9. These are also called the x-intercepts (if you were to graph the equation, you'd see it intersect the x-axis at these x-values).

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