Answer
$y= C_{1}e^{-t}+C_{2}e^{(-2-\sqrt{2}) t}+C_{3}e^{(-2+\sqrt{2}) t}$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''+5{y}''+6{y}'+2y=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}+5r^2e^{rt}+6re^{rt}+2e^{rt}=0\\\\$
$r^3+5r^2+6r+2=(r+1)(r^2+4r+2)=0 \;\;\;\;\;\;$$\rightarrow \;\;\;\;\;\; r_{1}= -1\;\;\;\;\;\;\;or\;\;\;\;\;\;r_{2}=-2-\sqrt{2}\;,\;r_{3}=-2+\sqrt{2} \;\;\;\;\;\;\\\\$
$y= C_{1}e^{-t}+C_{2}e^{(-2-\sqrt{2}) t}+C_{3}e^{(-2+\sqrt{2}) t}$
