There seems to be a problem with this question. The usual convention is that $\sqrt m$ refers only to the non-negative root, but in this case, the question is only solvable (even in the reals) if you allow for the negative root.
Let $\sqrt m = u$. The equation can be rewritten as $u^3 - u^2 + 80 = 0$, which can be shown (by sketching the curve for $y = u^3-u^2 = u^2(u-1)$ to guide the root search and the use of rational root theorem) to have a single real root at $u = -4$.
This leaves us in a bit of a quandary. If we adopted the convention we used above (take the negative root), the second expression works out to be $36$. But using the usual convention, we get $-4$. Perhaps both answers are meant to be stated.
By the way, the cubic also has a pair of conjugate complex roots, which weren't even considered in my answer. If you believe this is required, then you should work out the full solution of the cubic (not that hard to do once you've got the first factor of $u+4$ and divided by it to get a quadratic).