jmc

Rearranging the expression, we have $$x^2+2x+y^2-4y+8$$Completing the square in $x$, we need to add and subtract $(2/2{)}^2=1$. Completing the square in $y$, we need to add and subtract $(4/2{)}^2=4$. Thus, we have $$(x^2+2x+1)-1+(y^2-4y+4)-4+8\Rightarrow (x+1{)}^2+(y-2{)}^2+3$$Since the minimum value of $(x+1{)}^2$ and $(y-2{)}^2$ is $0$ (perfect squares can never be negative), the minimum value of the entire expression is $\boxed{3}$, and is achieved when $x=-1$ and $y=2$.