Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?

After writing our equation in the form $$\sqrt{(x+1)(x+7)}+\sqrt{(x+1)(x+2)}=\sqrt{(x+1)(6x+13)}$$ we obtain the domain:
$$(-\infty-7]\cup[-1,+\infty).$$
1. $x\geq-1.$
We see that $-1$ is a root and it remains to solve here 
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ or
$$\sqrt{(x+7)(x+2)}=2(x+1)$$ or
$$3x^2-x-10=0,$$ which gives also $x=2.$


*$x<-1$.


Thus, $x\leq-7$ and we need to solve
$$\sqrt{-x-7}+\sqrt{-x-2}=\sqrt{-6x-13}$$ or
$$\sqrt{(x+7)(x+2)}=-2(x+1),$$ which gives $$3x^2-x-10=0$$ again or
$$(x-2)(3x+5)=0$$ and we see that this equation has no roots for $x\leq-7.$
Id est, we got the answer: $$\{-1,2\}$$
