# How do you solve 5x^2 + 12r = -4x^2 -4?

$5 {x}^{2} + 12 r = - 4 {x}^{2} - 4$
$\Rightarrow 9 {x}^{2} = - 12 r - 4$
$\Rightarrow {\left(3 x\right)}^{2} = - 12 r - 4$
$\Rightarrow 3 x = \pm \sqrt{- 12 r - 4}$
$\Rightarrow x = \pm \frac{\sqrt{- 12 r - 4}}{3}$