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

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.

Let $y:=x-\dfrac{3}{2}$. The equation becomes $$\frac{1}{\left(y+\frac{3}{2}\right)^2}+\frac{1}{\left(y-\frac{3}{2}\right)^2}=\frac{104}{25}\,.$$ This is equivalent to $$\frac{y^2+\frac{9}{4}}{\left(y^2-\frac{9}{4}\right)^2}=\frac{52}{25}\,.$$ Let $z:=\dfrac{1}{y^2-\frac{9}{4}}$, we have $$\frac{9}{2}z^2+z=z^2\left(\frac{1}{z}+\frac{9}{2}\right)=\frac{52}{25}\,.$$ That is, $$\frac{9}{2}\left(z+\frac{4}{5}\right)\left(z-\frac{26}{45}\right)=0\,.$$ Thus, $z=-\dfrac45$ or $z=\dfrac{26}{45}$.

In the case $z=-\dfrac{4}{5}$, we have $$y^2-\frac{9}{4}=\frac{1}{z}=-\frac{5}{4}\,,$$ so $y^2=1$, or $y=\pm1$. In this case, $x=\dfrac{1}{2}$ or $x=\dfrac{5}{2}$.

In the case $z=\dfrac{26}{45}$, we have $$y^2-\frac{9}{4}=\frac{1}{z}=\frac{45}{26}\,.$$ That is, $y^2=\dfrac{207}{52}$, so $y=\pm\dfrac{3\sqrt{299}}{26}$. Hence, $$x=\frac{39\pm3\sqrt{299}}{26}\,.$$

Now you use the rational root theorem, Horner's algorithm and a little work to find the roots of the polynomial.

The rational candidates for the root are $\frac{p}{q}$ where $p$ is a factor of $225$ (i.e., $3^2\cdot 5^2$) and $q$ is a factor of $104$ (i.e., $2^3\cdot 13$).