Answer
Solution set = $\displaystyle \{\frac{\ln 3}{\ln 2}\}.$
$ x\approx 1.58$
Work Step by Step
Recognize that $2^{2x}=(2^{x})^{2}$. Introduce a new variable t:
$ t=2^{x}.$
$ t \gt 0 $, because $2^{x}$ is always positive.
We solve
$ t^{2}+t-12=0$
Factor by finding factors of $-12$ whose sum is $+1$:
$(t+4)(t-3)=0$
$ t+4=0$ or $ t-3=0$
$ t=-4$ or $ t=3$
$ t=-4$ is discarded, as $ t \gt 0 $, leaving
$ t=3\qquad $... bring back x
$ 2^{x}=3\qquad $... apply ln( )
$\ln 2^{x}=\ln 3\qquad $... apply $\log_{b}a^{p}=p\log_{b}a $
$ x\ln 2=\ln 3\qquad $... divide with $\ln 2$
$ x=\displaystyle \frac{\ln 3}{\ln 2}\qquad $... round to 2 decimal places
$ x\approx 1.58$