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.

Divide the whole equation by $x^2$ and simplifing:
$$6(x^2+\frac{1}{x^2})-25(x-\frac{1}{x})+12=6((x-\frac{1}{x})^2+2)-25(x-\frac{1}{x})+12=0$$
Put $x-\frac{1}{x}=y$ to solve the resulting equation:
$6y^2-25y+24=0$
