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 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!

Since $(y+1)^2\ge 0$ we must have $(x+y)^2+7(x+y)+10=(x+y+5)(x+y+2)\leq 0$ then, solving this inequality for $x+y$, we get $$-5\le x+y \le -2$$
