# How do solve x^2<10-3x and write the answer as a inequality and interval notation?

Jan 31, 2017

The answer is $- 5 < x < 2$
x in ]-5, 2[

#### Explanation:

Let's rewrite the inequality

${x}^{2} + 3 x - 10 < 0$

Let's factorise

$\left(x - 2\right) \left(x + 5\right) < 0$

Let $f \left(x\right) = \left(x - 2\right) \left(x + 5\right)$

We can build the sign chart

$\textcolor{w h i t e}{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}$$- 5$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\textcolor{w h i t e}{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}$$x - 2$$\textcolor{w h i t e}{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}$$f \left(x\right)$$\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}$$+$

Therefore,

$f \left(x\right) < 0$ when x in ]-5, 2[

or $- 5 < x < 2$