jmc

We try to factor the equation piece by piece. Start with the terms $2000x^6$ and $-2,$ and use difference of cubes: $$\begin{array}{rl}2000x^6-2& =2((10x^2{)}^3-1)\\ & =2(10x^2-1)(100x^4+10x^2+1)\\ & =(20x^2-2)(100x^4+10x^2+1).\end{array}$$Now we notice that the remaining terms make $$100x^5+10x^3+x=x(100x^4+10x^2+1),$$so we can factor the whole left-hand side, giving $$(20x^2+x-2)(100x^4+10x^2+1)=0.$$The term $100x^4+10x^2+1$ is always positive for real $x$, so the two real roots must be the roots of the quadratic $20x^2+x-2=0$. By the quadratic formula, $$x=\frac{-1\pm \sqrt{1^2+4\cdot 2\cdot 20}}{40}=\frac{-1\pm \sqrt{161}}{40}.$$The difference between these roots is $\frac{\sqrt{161}}{20}$, so the answer is $\boxed{\frac{161}{400}}$.