Answer
$\{1\}$
Work Step by Step
...$2^{x+2}=2^{x}\cdot 2^{2}=4(2^{x})$
... Substitute $t=2^{x}$
... the equation becomes
$t^{2}+4t-12=0 \quad $... factors of $-12$ whose sum is $+4$...
$(t-2)(t+6)=0$
t is either $-6$ or $2$.
$t=2^{x}=-6$ cannot be a solution as $2^{x}$ is never negative.
$t=2^{x}=2$
$2^{x}=2^{1}\quad$...Apply$: \quad $If $a^{u}=a^{v},$ then $u=v$
$x=1$
The solution set is $\{1\}$
