# How do you solve n^2+2n-24<=0?

Jan 28, 2017

The answer is $n \in \left[- 6 , 4\right]$

#### Explanation:

Let's factorise the inequality

${n}^{2} + 2 n - 24 = \left(n + 6\right) \left(n - 4\right)$

Let $f \left(n\right) = \left(n + 6\right) \left(n - 4\right)$

Now, we can build the sign chart

$\textcolor{w h i t e}{a a a a}$$n$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$n + 6$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$n - 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(n\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(n\right) \le 0$ when $n \in \left[- 6 , 4\right]$