Answer
$x=\frac{1}{2}$ or $x=2$ or $x=-2$
Work Step by Step
We solve by factoring with grouping:
$2x^{3}+4=x^{2}+8x$
$2x^{3}+4-x^{2}-8x=0$
$2x^{3}-x^{2}-8x+4=0$
$x^{2}(2x-1)-4(2x-1)=0$
$(2x-1)(x^{2}-4)=0$
$(2x-1)(x-2)(x+2)=0$
$2x-1=0$ or $x-2=0$ or $x+2=0$
$2x=1$ or $x=2$ or $x=-2$
$x=\frac{1}{2}$ or $x=2$ or $x=-2$