Answer
$x=0$
$x=-2-\sqrt{2}$
$x=-2+\sqrt{2}$
Work Step by Step
We factor out $x^2$:
$x^{4}+4x^{3}+2x^{2}=0$
$x^{2}(x^{2}+4x+2)=0$
$x^{2}=0$ or $(x^{2}+4x+2)=0$
Left solution: $x=0$
Right solution: use the quadratic formula
$x^{2}+4x+2=0$
$x=\displaystyle \frac{-4\pm\sqrt{4^{2}-4*1*2}}{2(1)}=\frac{-4\pm\sqrt{16-8}}{2}=\frac{-4\pm\sqrt{8}}{2}=\frac{-4\pm 2\sqrt{2}}{2}=-2\pm\sqrt{2}$
So the solutions are: $x=0$, $x=-2-\sqrt{2}$ and $x=-2+\sqrt{2}$
