Complete the squares:
[tex]x^2+2x=x^2+2x+1-1=(x+1)^2-1[/tex]
[tex]y^2-4y=y^2-4y+4-4=(y-2)^2-4[/tex]
So we can rewrite the given expression as
[tex]\bigg((x+1)^2-1\bigg)+\bigg((y-2)^2-4\bigg)+8=(x+1)^2+(y-2)^2+3[/tex]
Both [tex](x+1)^2[/tex] and [tex](y-2)^2[/tex] are non-negative for any choice of real [tex]x,y[/tex], which means the minimum value is 3 for [tex]x=-1[/tex] and [tex]y=2[/tex].
