CCSS Math: HSA.REI.B.4b
The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula.

## What is the quadratic formula?

$x=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}$
$\purpleD{a}x^2 + \goldD{b}x + \redD{c} = 0$

### Example

We're given an equation and asked to solve for $q$:
$0=-7q^2+2q+9$
This equation is already in the form $ax^2 + bx + c = 0$, so we can apply the quadratic formula where $a = -7, b = 2, c = 9$:
\begin{aligned} q &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ q &= \dfrac{-2 \pm \sqrt{2^{2} - 4 (-7) (9)}}{2(-7)} \\\\ q &= \dfrac{-2 \pm \sqrt{4 +252}}{-14} \\\\ q &= \dfrac{-2 \pm \sqrt{256}}{-14} \\\\ q &= \dfrac{-2 \pm 16}{-14} \\\\ q &= \dfrac{-2 + 16}{-14} ~~,~~ q = \dfrac{-2 - 16}{-14} \\\\ q &= -1 ~~~~~~~~~~~~,~~ q = \dfrac{9}{7} \end{aligned}
Let's check both solutions to be sure it worked:
$q=-1$$q=\dfrac{9}{7}$
\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7(-1)^2+2(-1)+9 \\\\0&=-7(1)-2+9 \\\\0&=-7-2+9\\\\0&=0\end{aligned}\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7\left(\dfrac{9}{7}\right)^2+2\left (\dfrac{9}{7}\right)+9 \\\\0&=-7\left(\dfrac{81}{49}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{81}{7}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{63}{7}\right) +9 \\\\0&=-9 +9 \\\\0&=0\end{aligned}
Yep, both solutions check out.
Solve for $x$.
$-4+x+7x^2 = 0$