2^(2x)-12*2^x+32=0 Solve the equation.

We are asked to solve 2^(2x)-12*2^x+32=0 :
Rewrite as (2^x)^2-12*2^x+32=0 and let y=2^x ; then
y^2-12y+32=0 and
(y-8)(y-4)=0 so y=8 or y=4.
y=8 ==> 2^x=8 ==> x=3
y=4 ==> 2^x=4 ==> x=2
Checking we find that these are indeed solutions.
The graph: