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,
[tex]2x^2-3x-10=-3x^2+20[/tex]
So, by moving the term -3x^3+20 to the left hand side, we have
[tex]5x^2-3x-30=0[/tex]
Then, in order to solve this equation, we can apply the quadratic formula
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In our case, a=5, b=-3 and c=-30. So we get
[tex]x=\frac{3\pm\sqrt{(-3)^2-4(5)(-30)}}{2(5)}[/tex]
which gives
[tex]\begin{gathered} x=2.76779 \\ and \\ x=-2.16779 \end{gathered}[/tex]
By substituting these points into one of the functions, we have
[tex]f(2.76779)=-2.982[/tex]
and
[tex]f(-2.16779)=5.902[/tex]
Then, by rounding these numbers to the nearest tenth, we have the following points:
[tex]\begin{gathered} (2.8,-3.0) \\ and \\ (-2.2,5.9) \end{gathered}[/tex]
Therefore, the answer is the last option
