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.

Substitute $y:=x-\frac{1}{x}$. Thus $y^2=x^2-2+\frac{1}{x^2}$.
Your original equation can be written as $6y^2-25y+24=0$.
Thus for y we get $\frac{25\pm\sqrt{625-96}}{12}=\frac{25\pm23}{12}=\{\frac{1}{6};4\}$
For $y=\frac{1}{6}$ we get for $x^2-\frac{1}{6}x+1=0$, which hasn't any real solution.
For $y=4$ we get $x^2-4x+1=0$, which has the solutions $\frac{4\pm\sqrt{16-4}}{2}=\underline{\underline{2\pm\sqrt{3}}}$.
