Answer: \(x=-5,\ \frac32,\ 2\).

Answer: \(x=-5,\ \frac32,\ 2\).

At the points of intersection, the two expressions for \(y\) are equal:

\(2x^3+3x^2-26x+22=3x-8.\)

Rearranging gives

\(2x^3+3x^2-29x+30=0.\)

Testing simple integer roots gives \(x=2\):

\(2(2)^3+3(2)^2-29(2)+30=0.\)

So \((x-2)\) is a factor. Dividing gives

\(2x^3+3x^2-29x+30=(x-2)(2x^2+7x-15).\)

Now factorise the quadratic:

\(2x^2+7x-15=(2x-3)(x+5).\)

Therefore

\((x-2)(2x-3)(x+5)=0.\)

The required \(x\)-coordinates are

\(x=2,\quad x=\frac32,\quad x=-5.\)