Answer
$x=8$
Work Step by Step
We are given:
$\sqrt{2x}=\sqrt{3x+12}-2$
We square both sides:
$(\sqrt{2x})^{2}=(\sqrt{3x+12}-2)^{2}$
$(\sqrt{2x})(\sqrt{2x})=(\sqrt{3x+12}-2)(\sqrt{3x+12}-2)$
$2x=3x+12-4\sqrt{3x+12}+4$
$2x-3x-16=-4\sqrt{3x+12}$
$-x-16=-4\sqrt{3x+12}$
$4\sqrt{3x+12}=x+16$
We square both sides again:
$(4\sqrt{3x+12})^{2}=(x+16)^{2}$
$16\ (3x+12)=x^{2}+32x+256$
$48x+192=x^{2}+32x+256$
$x^{2}-16x+64=0$
And factor:
$(x-8)^{2}=0$
Use the zero-factor property by equating each factor to zero:
$x-8=0$
$x=8$