Answer
$-4, -2$
Work Step by Step
$y=e^{rx}$
$y'=e^{rx}\frac{d}{dx}(rx)$
$y'=re^{rx}$
$y''=re^{rx}\frac{d}{dx}(rx)$
$y''=r^2e^{rx}$
$y''+6y'+8y=0$
$r^2e^{rx}+re^{rx}+8e^{rx}=0$
$r^2+r+8=0$
$(r+4)(r+2)=0$
So $r=-4$ and $r=-2$ are the solutions.