The answer is “no”. Note that $$\begin{array}{rcl}& & z^2+(-2+2i)z-(1+2i)=0\\ & \iff & z^2+(-2+2i)z=1+2i\\ & \iff & z^2+(-2+2i)z+{\left(\frac{-2+2i}{2}\right)}^2=1+2i+{\left(\frac{-2+2i}{2}\right)}^2\\ & \iff & {(z+\frac{-2+2i}{2})}^2=1+2i+(-1+i{)}^2\\ & \iff & (z+(-1+i){)}^2=1+2i-2i\\ & \iff & (z+(-1+i){)}^2=1\end{array}$$ Since the square roots of $1$ are $1$ and $-1$, we must have $z=1-(-1+i)=2-i$ or $z=-1-(-1+i)=-i$. In summary, the solutions are $2-i$ and $-i$ and neither is a real number.