Answer
$(-\infty,-8)\cup(-6,4)\cup(6,\infty)$
Work Step by Step
Step 1. The given inequality can be separated into two cases:
$x^2+2x-36\gt12$ and $x^2+2x-36\lt-12$
Step 2. For the first case, we have
$x^2+2x-48\gt0$, $(x+8)(x-6)\gt0$
and the boundary points are $x=-8,6$
Using the test points to examine signs on the left side across the boundary points, we have
$...(+)...(-8)...(-)...(6)...(+)...$
Thus the solutions are
$x\lt-8$ or $x\gt6$; that is, $(-\infty,-8)\cup(6,\infty)$
Step 3. For the second case, we have
$x^2+2x-24\lt0$, $(x+6)(x-4)\gt0$
and the boundary points are $x=-6,4$.
Using test points to examine signs of the left side across the boundary points, we have
$...(+)...(-6)...(-)...(4)...(+)...$
Thus the solutions are
$-6\lt x\lt 4$; that is, $(-6,4)$
Step 4. Combining the above results, we have
$(-\infty,-8)\cup(-6,4)\cup(6,\infty)$
Step 5. The results are shown on a real number line in the figure.
