#### Answer

$\left\{-2, 4\right\}$

#### Work Step by Step

RECALL:
A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$.
The trinomial's factored form will be:
$x^2+bx+c=(x+d)(x+e)$
The given trinomial has $b=2$ and $c=-8$.
Note that $-8=-4(2)$ and $2= -4+2$.
This means that $d=-4$ and $e=2$
Thus, the factored form of the trinomial is:
$[x+(-4)](x+2) = (x-4)(x+2)$
The given equation maybe written as:
$(x-4)(x+2)=0$
Use the Zero-Factor Property by equating each factor to zero.
Then, solve each equation to obtain:
$\begin{array}{ccc}
&x-4 = 0 &\text{ or } &x+2=0
\\&x=4 &\text{ or } &x=-2
\end{array}$
Thus, the solution set is $\left\{-2, 4\right\}$.