The solution of the system of equation is the intersection point of the two quadratic equations, so we need to equate both equations, that is,

So, by moving the term -3x^3+20 to the left hand side, we have

Then, in order to solve this equation, we can apply the quadratic formula

In our case, a=5, b=-3 and c=-30. So we get

which gives

By substituting these points into one of the functions, we have

and

Then, by rounding these numbers to the nearest tenth, we have the following points:

Therefore, the answer is the last option