We try to factor the equation piece by piece. Start with the terms $ 2000x^6 $ and $ -2, $ and use difference of cubes: \[\begin{aligned} 2000x^6 - 2 & = 2((10x^2)^3 - 1) \\ &= 2(10x^2-1)(100x^4 + 10x^2 + 1) \\ &= (20x^2-2)(100x^4+10x^2+1).\end{aligned}\]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}} $.