This problem can be solved using geometric method. The given equation can be rewritten as $$y=\sqrt{(x+1)^2+(0-2)^2}+\sqrt{(x-2)^2 +(0+1)^2}$$ Therefore, finding the minimal value of $y$ is equivalent to find a point on $x-$axis such that the sum of its distance to point $A (-1, 2)$ and point $B (2, -1)$ minimizes. Notice that $A$ and $B$ are on different sides of the $x-$axis. Therefore, the minimal value of the sum is simply the distance between $A$ and $B$ which is $$\sqrt{(-1-2)^2 + (2-(-1)^2)}=\boxed{3\sqrt{2}}$$