Algebra

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! :) :) :)