Answer
$(-\infty,-2)U(3,\infty)$
Work Step by Step
Step 1. Factor the quadratic, we have $(x+2)(x-3)\gt0$
Step 2. The boundary points are $x=-2$ and $x=3$
Step 3. Use test points at $x=-3, 0, 4$, the signs of the quadratic are $+, -, +$
Step 4. The inequality requires positive, thus the solutions are $(-\infty,-2)U(3,\infty)$