I did $-(x+2)^2+6(x+2)>13$ and $-(x+2)^2+6(x+2)< -13$. The first inequality had complex solutions and therefore can be disregarded but the second one has two real solutions, $x \approx -3.7$ and $5.7$.
So because the graph is negative, we know that to the left and to the right of these two values, all of the $y$ values are $< -13$ so therefore the solution in interval notation is $(-\infty, -3.7) \cup (5.7, \infty)$. I'm not sure my logic is correct in the last part of this. Thank you.
Let $y = x+2$. Then the inequality becomes $|-y^2+6y| > 13$, which you asked here .
Now, just shift all of the intervals to that answer by $2$ to get the answer to this problem.