How would I go about solve for x in this equation:
$\displaystyle 2^{(2x+2)}-17\times2^x+4=0$
How would I go about solving it since 17 isn't a power of 2 and I can't figure out a way to take 17?
$\displaystyle 2^{(2x+2)}-17\times2^x+4=0$
$\displaystyle 2^{2}\times 2^{2x}-17\times2^x+4=0$
$\displaystyle 4\times 2^{2x}-17\times2^x+4=0$
make $\displaystyle a = 2^x$
$\displaystyle 4 a^2-17a+4=0$
Now you have a quadratic. Can you you solve it from here?