# How do you solve 2x=-6x^2?

$x$= $0$ or $- \frac{1}{3}$
$2 x$ = $- 6 {x}^{2}$
$6 {x}^{2} + 2 x = 0$
$2 x \left(3 x + 1\right) = 0$
$2 x = 0$ or $\left(3 x + 1\right) = 0$
$x$= $0$ or $- \frac{1}{3}$