## Statement

Given the equation:

$$ax^2 + bx + c = 0$$

The values of $x$ can be found using this formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

## Proof

$$ax^2 + bx + c = 0$$

Divide every term by $a$.

$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

Complete the square by adding and subtracting $\left(\frac{b}{2a}\right)^2$.

$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$$

$$\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$$

Isolate the square.

$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$

Expand the brackets on the right hand side and combine the fractions.

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}$$

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$

Take the square root on both sides. Note that this will result in a $\pm$-sign on the right hand side.

$$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$

Now isolate the $x$-term.

$$x = - \frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$

Finally, simplify the result.

$$x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$$

$$x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

## Proofs building upon this proof

### Trig values of 18 degrees

This proof shows the exact value of sin(18), cos(18) and tan(18)