Answer
$x=-4$ and $x=2$
Work Step by Step
$x^{2}e^{2x}+2xe^{2x}=8e^{2x}$
Take $8e^{2x}$ to the left side:
$x^{2}e^{2x}+2xe^{2x}-8e^{2x}=0$
Take out common factor $e^{2x}$ from the left side:
$e^{2x}(x^{2}+2x-8)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$e^{2x}=0$
There are no values of $x$ that make this equation true.
$x^{2}+2x-8=0$
Solve by factoring:
$(x+4)(x-2)=0$
$x+4=0$
$x=-4$
$x-2=0$
$x=2$
The solutions found are $x=-4$ and $x=2$
