Intersection points, part 2

I can only give a partly solution because the way i solved it is very tedious and i dont know if there is a trick that makes the solving process easy and intuitive. I will first give an intuition to my solution:

If $f$ and $f^{-1}$ intersect in point $(a,b)$ it must mean that point $(b,a)$ is also a point of intersection. For $f^{-1}$ and every input $x$ the following must apply - $f^{-1}(f(x)) =x$ . Therefore on the intersection points the following must also apply - $f(f(x)) =x$ . This is only true if $x$ is an intersection point!

So this is what i did. I mounted $f$ on $f$ and i want to look for all values $x$ that hold $f(f(x)) =x$ . For convenience i denoted $f=\frac{-1}{10}g$ where $g$ is the polynomial. After a little algebric work we get to the following equation:

$1000(10x-18)=g^3-30g^2+400g$ . Now all we have to do is insert the polynomial instead of $g$ , open up all the brackets and convene everything into one big polynomial. After some very tedious algebric work (which i used wolfram alpha for) we get the following equation:

$x^9+9x^8+39x^7+15x^6-348x^5-1284x^4+500x^3+6012x^2-192x-4752=0$ . This polynomial has 5 real roots and 4 complex roots. We are interested only in the real roots. With the help of wolfram alpha again, i extracted the real roots of the polynomial and they are $-1,1,2,-\sqrt{6},\sqrt{6}$ and these are all the values $a_i$ . Therefore $S= 4+2\sqrt{6}=8.89897...$ . And the answer is $\lfloor 1000S \rfloor = \boxed{8898}$ .