Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?

Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.


$$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$
$$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$
$$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$
$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$
