Answer
$(-\infty,-7)\cup(1,\infty)$
Work Step by Step
Step 1. The given inequality can be separated into two cases:
$x^2+6x+1\gt8$ and $x^2+6x+1\lt-8$
Step 2. For the first case, we have
$x^2+6x-7\gt0$, $(x+7)(x-1)\gt0$
and the boundary points are $x=-7,1$
Using the test points to examine signs of the left side across the boundary points, we have
$...(+)...(-7)...(-)...(1)...(+)...$
Thus the solutions are
$x\lt-7$ or $x\gt1$, that is, $(-\infty,-7)\cup(1,\infty)$
Step 3. For the second case, we have
$x^2+6x+9\lt0$, $(x+3)^2\lt0$
which has no solution.
Step 4. Combining the above results, we have
$(-\infty,-7)\cup(1,\infty)$
Step 5. The results are shown on a real number line in the figure.
