$\displaystyle 4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2$

This is as far as i got:

$\displaystyle \frac{1}{2}^{-2x}*\frac{1}{2}^{3-2x}=\frac{1}{2}^{-3}*\frac{1}{2}^{-x^2}$

$\displaystyle 2x*3-2x=3*-x^2$

$\displaystyle 6x-2x=-3x^2$

$\displaystyle 3x^2+4x=0$

any help would be much appriciated!