smc

To solve this problem, we may begin by factoring $x^2-x-6$ as $(x-3)(x+2)$. This is an upward opening parabola, therefore the solutions to $(x-3)(x+2)<0$ are between the roots of the equation. That means our solutions are all $x$ such that $-2<x<3$, or simply $\boxed{-2<x<3}$.