Solving $\;5^{2x}-4\cdot 5^x=12$ I need to solve $\quad\displaystyle 5^{2x}-4\cdot 5^x=12$.
I've only gotten this far: $\quad \displaystyle 5^{2x}-20^x=12.$
I don't know what to do next.
Thanks in advance!

Please note that $4\cdot5^x = 4(5^x)\neq 20^x$ 
If we let $y=5^x$, then $$(5^x)^2-4(5^x) =12 \;\implies \;y^2 - 4y = 12 \;\implies y^2 - 4y - 12 = (y-6)(y+2) = 0$$
$\implies \; y = 6\;$ or $\;y = -2$
So $y = 5^x = 6\; $ or $\; y = 5^x = -2$
Can you take it from here? 
You can omit $\;y= 5^x = -2\;$ as a solution if we are constraining ourselves to real solutions...
So it suffices to solve for $x$ given $\;5^x = 6$.
