$20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$, then $x$ is equal to :

Question

( A ) $\pm \frac{\sqrt{4}}{3}$
( B ) $\pm \frac{\sqrt{3}}{2}$
( C ) $\pm \frac{\sqrt{2}}{3}$
( D ) $\pm \frac{\sqrt{5}}{4}$