Answer: \(\left(-\dfrac1{20},-\dfrac{31}{5}\right)\).

At the intersections,

\(10x^3-19x^2-x=4x-6.\)

Bring all terms to one side:

\(10x^3-19x^2-5x+6=0.\)

Since \(C=(2,2)\), \(x=2\) is one root. Therefore \(x-2\) is a factor.

Dividing gives

\(10x^3-19x^2-5x+6=(x-2)(10x^2+x-3).\)

Now factorise the quadratic:

\(10x^2+x-3=(5x+3)(2x-1).\)

So the other two \(x\)-coordinates are

\(x=-\frac35,\qquad x=\frac12.\)

These correspond to \(A\) and \(B\). Use \(y=4x-6\) to find their \(y\)-coordinates:

\(x=-\frac35\quad\Rightarrow\quad y=4\left(-\frac35\right)-6=-\frac{42}{5},\)

and

\(x=\frac12\quad\Rightarrow\quad y=4\left(\frac12\right)-6=-4.\)

The midpoint of \(AB\) is therefore

\(\left(\frac{-\frac35+\frac12}{2},\frac{-\frac{42}{5}-4}{2}\right).\)

This simplifies to

\(\boxed{\left(-\frac1{20},-\frac{31}{5}\right)}.\)