Answer
$x=\displaystyle \frac{5}{3}$
Work Step by Step
We solve by factoring with grouping:
$3x^{3}+12x=5x^{2}+20$
$3x^{3}+12x-5x^{2}-20=0$
$3x^{3}-5x^{2}+12x-20=0$
$x^{2}(3x-5)+4(3x-5)=0$
$(3x-5)(x^{2}+4)=0$
$3x-5=0$ or $x^{2}+4=0$
$3x=5$ or $x^{2}=-4$
$x=\displaystyle \frac{5}{3}$ or $x=\pm \sqrt{-4}=$ not real
$x=\displaystyle \frac{5}{3}$