# How do you solve 6x^2+3x>= -5x-3+x^2?

Jan 21, 2017

The answer is x in ]-oo,-1]uu [-3/5, +oo[

#### Explanation:

Let'e rearrange the inequality

$6 {x}^{2} + 3 x \ge - 5 x - 3 + {x}^{2}$

$5 {x}^{2} + 8 x + 3 \ge 0$

We factorise

$\left(5 x + 3\right) \left(x + 1\right) \ge 0$

Let $f \left(x\right) = \left(5 x + 3\right) \left(x + 1\right)$

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}$$- 1$$\textcolor{w h i t e}{a a a a}$$- \frac{3}{5}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\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}$$5 x + 3$$\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 a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x \ge 0\right)$ when x in ]-oo,-1]uu [-3/5, +oo[