# How do you solve x^2 - 2x - 18 = 0 using the quadratic formula?

Aug 6, 2015

$x = 1 + \sqrt{19}$ or $x = 1 - \sqrt{19}$
(see below for solution using the quadratic formula)

#### Explanation:

$\textcolor{w h i t e}{\text{XXXX}}$$a {x}^{2} + b x + c = 0$
the roots (solution) can be determined by:
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the given equation ${x}^{2} - 2 x - 18 = 0$
$a = 1$$\textcolor{w h i t e}{\text{XXXX}}$$b = - 2$$\textcolor{w h i t e}{\text{XXXX}}$$c = - 18$

So the solutions are
$\textcolor{w h i t e}{\text{XXXX}}$$x = \frac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(1\right) \left(- 18\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{2 \pm \sqrt{4 + 72}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{2 \pm \sqrt{76}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{2 \pm 2 \sqrt{19}}{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 1 \pm \sqrt{19}$