The number of real roots of  $\sqrt{x^2-x-6}+\sqrt{6x^2-5x-39}=\sqrt{x^2+4x-21}$  is 

The given equation is $\sqrt{6x^2-5x-39}=\sqrt{x^2+4x-21}-\sqrt{x^2-x-6}$

This can be written as 

$\sqrt{\left(6x+13\right)\left(x-3\right)}=\sqrt{\left(x+7\right)\left(x-3\right)}-\sqrt{\left(x-3\right)\left(x+2\right)}$

take common $\sqrt{x-3}$ from each term and equate to zero, $x=3$, is one root to the given equation.

The remaining equation is 

$\sqrt{6x+13}=\sqrt{x+7}-\sqrt{x+2}$

Squaring on both sides

$\begin{array}{rcl}6x+13& =& x+7+x+2-2\sqrt{x+7}\sqrt{x+2}\\ 4x+4& =& -2\sqrt{x^2+9x+14}\\ 2x+2& =& -\sqrt{x^2+9x+14}\end{array}$

Squaring on both sides

$\begin{array}{rcl}4x^2+8x+4& =& x^2+9x+14\\ 3x^2-x-10& =& 0\\ 3x^2-6x+5x-10& =& 0\\ 3x\left(x-2\right)+5\left(x-2\right)& =& 0\\ \left(3x+5\right)\left(x-2\right)& =& 0\end{array}$

$x=-\frac{5}{3},x=2$ are extrenious solutions becasue these two or not in domian

Hence, the number of solutions for the given equation is 1. 

S.O.B.S, twice and solving we get only 2 real roots