How do you solve 5x<=2-3x^2?

Oct 16, 2016

$- 2 \le x \le \frac{1}{3}$

Explanation:

$5 x \le 2 - 3 {x}^{2}$
Rearranging
$3 {x}^{2} + 5 x - 2 \le 0$
$\left(3 x - 1\right) \left(x + 2\right) \le 0$
$x = \frac{1}{3} \mathmr{and} x = - 2$
$- 2 \le x \le \frac{1}{3}$
Checking, let $x = - 1$
$- 4 \cdot 1 = - 4$ which $\le 0$