## Calculus 8th Edition

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