Answer
$(-\infty,-4)\cup(3,\infty)$ or $\{x|x\lt-4\ or\ x\gt3\}$.
Work Step by Step
Step 1. $x^2+x\gt12 \longrightarrow x^2+x-12\gt0 \longrightarrow (x+4)(x-3)\gt0$.
Step 2. Identify boundary points $x=-4,3$ and form intervals $(-\infty,-4),(-4,3),(3,\infty)$.
Step 3. Choose test values $x=-5,0,4$ to test the inequality and get results $True,\ False,\ True$.
Step 4. Thus we have the solution $(-\infty,-4)\cup(3,\infty)$ or $\{x|x\lt-4\ or\ x\gt3\}$.
