ّ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.

Let $u=x$ and $v=3-x$. Then we have
$$
u+v = 3,
\qquad
\frac{1}{u^2} + \frac{1}{v^2} = \frac{104}{25}
$$
But
$$
\frac{1}{u^2} + \frac{1}{v^2}
=
\frac{u^2+v^2}{(uv)^2}
=
\frac{(u+v)^2-2uv}{(uv)^2}
=
\frac{9-2uv}{(uv)^2}
$$
Thus, $uv$ is a root of
$$
\frac{9-2z}{z^2} = \frac{104}{25}
$$
a quadratic equation. Once you know $uv$ and $u+v$, you know $u$ and $v$ by solving another quadratic equation.
