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!

Set $x+y=c\iff y=c-x$
We have $(c+1)^2+5c+2y+10+y^2=0\iff y^2+2y+c^2+7c+11=0$
As $y$ is real, we need the discriminant $(2)^2-4\cdot1\cdot(c^2+7c+11)\ge0$
$\iff(c+2)(c+5)\le0$
