Given is the following differential equation $$x'=y\\ y'=-2x-4\epsilon x^3 $$ with $\epsilon \geq 0$.
I was asked to calculate the fixed points, so I found that $y=0$ and for $-2x-4\epsilon x^3=0$, I found $x=0$ or $x=\pm \sqrt\frac{1}{2\epsilon} i$.
I know that $(0,0)$ is a fixed point, but I don't know what to do with these complex solutions for $x$. Any help would be appreciated!