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?

The usual way to do something like this is to square to get rid of one radical, rearrange, and square again to get rid of the remaining radical, as @vhspdfg says in a comment. I won’t detail the intermediate steps, but the octic polynomial I got was
$$
 2000 - 1600x - 3480x^2 + 2960x^3 + 825x^4 - 1340x^5 + 510x^6 - 140x^7 + 25x^8\\
=5(x+1)^2(x-2)^3(5x^3  - 8x^2 + 65x - 50)\,.
$$
The cubic seems to have no rational roots, and if this is correct, it’s irreducible.
