Answer
$x=\left\{-4, 2\right\}$
Work Step by Step
Subtract $8e^{2x}$ on both sides of the equation to obtain:
$x^2e^{2x} + 2xe^{2x}-8e^{2x} = 0$
Factor the trinomial to obtain:
$e^{2x}(x+4)(x-2)=0$
RECALL:
$a\cdot b \cdot c = 0 \longrightarrow a = 0 \text{ or } b=0 \text{ or } c=0$
Use the rule above by equating each factor to zero then solving each equation to obtain:
$\begin{array}{ccccc}
&e^{2x}=0 &\text{ or } &x+4=0 &\text{ or } &x-2=0
\\& & &x=-4 &\text{ or } &x=2
\end{array}$
Note that $e^{2x}=0$ has no real number solution as there is no real number that will make $e^{2x}=0$.
Thus, the solutions are $x=\left\{-4, 2\right\}$.
