# How do you solve x^2 - 3x - 18 = 0?

Nov 2, 2015

${x}_{1} = 6$ and ${x}_{2} = - 3$

#### Explanation:

you have an Equation like this: $a {x}^{2} + b x + c = 0$

Then you can use the formula for solving quadratic Equations:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

so we have: $x = - \left(- 3\right) \left(\pm\right) \frac{\sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 18\right)}}{2 \cdot 1}$

$x = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}$

${x}_{1} = \frac{3 + 9}{2} = 6$
${x}_{2} = \frac{3 - 9}{2} = - 3$