How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I see that, the equation
$$\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2} = 0$$
has unique solution $x = 2$, but I can not solve. How can I solve this equation or solve the given equation with another way?

First of all note that the RHS of your original equation can be written as $4(x-2)+3(x+1)$. Now transfer the terms on either side of the equation obtaining $$2(x-2)\big[\sqrt {5-x^2}-2\big] = (x+1)\big[3-\sqrt{5+x^2}\big].$$ The LHS vanishes for $\pm 1$ and $2$. The RHS vanishes for $-1$ and $\pm2$. Two of the  root are therefore $-1$ and $2$. 
EDIT: Since the RHS remains positive and the LHS remains negative from $-1$ to $2$, there are no further roots between these two. In the range $-\sqrt 5$ to -1, the LHS is more than RHS and in the range $2$ to $\sqrt 5$, LHS is less than RHS
