Example 3 (Question from the 5th "Hope Cup" Invitational Competition) The solution set of the inequality $\sqrt{2 x+2}>x-1$ is $\qquad$
Solution: Fill in $[-1,2+\sqrt{5}]$. Reason: Solving the equation $\sqrt{2 x+2}=x-1$, we get $x=$ $2+\sqrt{5}$. As shown in Figure 5-3, considering the set of x-coordinates of points on the semiparabola $y=\sqrt{2 x+2}$ that lie above the line $y=$ $x-1$, we have $-1 \leqslant x<2+\sqrt{5}$.