# How do you solve the inequality x^2-3x-18>0?

Nov 5, 2016

The answer is $\left(- \infty < x < - 3\right)$$\cup$$\left(6 < x < + \infty\right)$

#### Explanation:

We factorise the expression ${x}^{2} - 3 x - 18 = \left(x + 3\right) \left(x - 6\right)$
then we do a sign chart
$\textcolor{w h i t e}{a a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$6$$\textcolor{w h i t e}{a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$+$
$\textcolor{w h i t e}{a a a a a}$$x - 6$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$+$
$\textcolor{w h i t e}{a}$(x-6((x+3)$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a}$$+$

So ${x}^{2} - 3 x - 18 > 0$ when

$- \infty < x < - 3$ and $6 < x < + \infty$