ّ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.

One can see that $1/2$ is a solution of $\frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. Then it is easy to see that also $5/2$ is a solution.
Can you proceed ?
