I have a problem:

> 
> For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that
>  $$-5 \le x+y \le -2.$$
> 
> 
> 

I have tried: 

I write $(x+y+1)^2+5x+7y+10+y^2=(x+y)^2+7(x+y)+(y+1)^2+10=0.$ 

Now I'm stuck :( 

Any help will be appreciated! Thanks!