How would I go about solving this?
You don't. That is not a "one to one" function and so does not have an inverse. If $\displaystyle f^{-1}(x)$ is the inverse of f(x), then we must have $\displaystyle f^{-1}(f(x))= x$, that is, $\displaystyle f^{-1}$ must 'undo' whatever f does. But f(6)= 0 and f(-1)= 0. $\displaystyle f^{-1}(0)$ can't be both 6 and -1!
Luke774, for some reason, chose to take only the "+" sign on the square root. That's valid but gives the inverse of a slightly different function: $\displaystyle f(x)= x^2- 5x- 6$ with $\displaystyle x\ge 5/2$ and undefined for x< 5/2. Same formula but different domain so different function.