A text-based proof (not video) of the quadratic formula.
$x=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}$
$\purpleD{a}x^2 + \goldD{b}x + \redD{c} = 0$
If you've never seen this formula proven before, you might like to watch a video proof, but if you're just reviewing or prefer a text-based proof, here it is:

## The proof

We'll start with the general form of the equation and do a whole bunch of algebra to solve for $x$. At the heart of the proof is the technique called $\blueD{\text{completing the square}}$. If you're unfamiliar with this technique, you may want to brush up by watching a video.

### Part 1: Completing the square

\begin{aligned} \purpleD{a}x^2 + \goldD{b}x + \redD{c} &= 0&(1)\\\\ ax^2+bx&=-c&(2)\\\\ x^2+\dfrac{b}{a}x&=-\dfrac{c}{a}&(3)\\\\ \blueD{x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(4)\\\\ \blueD{\left (x+\dfrac{b}{2a}\right )^2}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(5) \end{aligned}

## Part 2: Algebra! Algebra! Algebra!

Remember, our goal is to solve for $x$.
\begin{aligned} \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{c}{a}&(5) \\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2} &(6)\\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2-4ac}{4a^2}&(7)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}&(8)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(9)\\\\ x&=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(10)\\\\ x&=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}&(11) \end{aligned}
Ecco fatto!