Group them in a nice way

We may factor the equation as:

$\begin{aligned} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{aligned}$

Now $100x^4+10x^2+1\ge 1>0$ for real $x$ . Thus the real roots must be the roots of the equation $20x^2+x-2=0$ . By the quadratic formula the roots of this are:

$x=\dfrac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \dfrac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.$ Thus $r=\dfrac{-1+\sqrt{161}}{40}$ , and so the final answer is $-1+161+40 = \boxed{200}$ .