ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. 
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$

My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.

Now, you can make the following.
Easy to see that $\frac{1}{2}$ and $\frac{5}{2}$ they are roots of the equation,
which gives a factor $$(2x-1)(2x-5)=4x^2-12x+5$$ and
$$104x^4-624x^3+886x^2+150x-225=$$
$$=104x^2-312x^3+130x^2-312x^3+936x^2-390x-180x^2+540x-225=$$
$$=26x^2(4x^2-12x+5)-78(4x^2-12x+5)-45(4x^2-12x+5)=$$
$$=(4x^2-12x+5)(26x^2-78x-45),$$ which gives the answer:
$$\left\{\frac{1}{2},\frac{5}{2},\frac{3}{2}\left(1+\sqrt{\frac{23}{13}}\right),\frac{3}{2}\left(1-\sqrt{\frac{23}{13}}\right)\right\}$$
