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?

When you squared you forget to put the coefficient of 2. 
You should have $2\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$
Or $\sqrt{(x^2+8x+7)(x^2+3x+2)}=2(x+1)^2$
....
But anyway....
Square again.  But it'll help to factor components.
$(x^2+8x+7)(x^2+3x+2)=4(x+1)^4$
$(x+7)(x+1)(x+2)(x+1)=4(x+1)^4$.
If $x = -1$ we get $0=0$ which could be a solution although we fatored so late in the game we should check if that's extraneous.
Nope... way back in the beginning $x = -1$ would have given us $\sqrt 0 + \sqrt 0 =\sqrt 0$ which is fine.
SO $x =-1$ is a solution.  Divide both sides by $(x+1)^2$ and we tet
$(x+7)(x+2)= 4(x+1)^2$ which is just a quadratic...
$x^2 + 9x + 14 = 4x^2 + 8x + 4$ so
$3x^2 -x -10=0$ which we can solve with the quadratic equation.  (but well have to search for extraneous solutions.)
$x = \frac {1\pm \sqrt{1 + 120}}6 = \frac {1\pm 11}6= 2, -\frac 53$.
But $x =-\frac 53$ give $\sqrt {\frac {25}9 - \frac {40}3+7}= \sqrt{-\frac {32}9}$ which is not real.
So $x=2$ or $x=-1$
