Alright! I geuss I can try to solve this question!

First, let's note that the only points where the lines intersect is when we have \(f(x) = g(x)\).

This means that we only have to solve the equation \(x^3 + x^2 -3x + 5 = x^3 + 2x^2 \) to find the points where the lines intersect.

Combining like terms and moving all the terms to one side, we get the equation \(x^2 + 3x - 5 = 0\).

We can't factor this polynomial with integers, so the next best thing to do is to complete the square and find x that way.

We set up the equation by moving 5 to the other side, getting us \(x^2 + 3x = 5\). Now we complete the square by adding 9/4 to both sides.

\(x^2 + 3x + 9/4 = 5 + 9/4 \)

\((x + 3/2)^2 = 29/4\)

Now we just have to square root both sides to get us \(x+3/2= \pm \sqrt{29/4}\)

This means that \(x = \frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}\).

Subsituting these x values back into the g(x) and f(x) functions, we get \(y = \frac{-\sqrt{29}-3} 2, \frac{\sqrt{29}-3}{2}\)

So for the final answer, we have \((\frac{\sqrt{29}-3}{2}, \frac{-\sqrt{29}-3}{2}), (-\frac{\sqrt{29}-3}{2}, \frac{\sqrt{29}-3}{2})\)

Thanks! :) :) :)