<s>[INST] Solve for $x$ in $6x^2-25x+12+\frac{25}{x}+\frac{6}{x^2}=0$ I simplified to get $6x^4-25x^3+12x^2+25x+6=0$
Or $6 (x^2+1)^2+25x(1-x^2)=0$
Then I substituted $z=x^2+1$, to get $6z^2+25\sqrt{(z-1)}(2-z)=0$
I can't find a way to proceed further.
 [/INST] Just use the rational root theorem after you made it into a polynomial
Candidate roots: $\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{1}{6}$.
Testing them you find that al four roots are rational:
$x\in\{2,3,-\frac{1}{2},-\frac{1}{3}\}$
</s>