Answer
$(4,32), \quad \left(-\frac{14}{3}, \frac{2}{9}\right) $
Work Step by Step
Given $$\begin{cases}
y=2 x^2+5 x-20 \\
y=-x^2+3 x+36.
\end{cases}$$ Set the two functions equal and solve for $x$.
\begin{equation}
\begin{aligned}
2 x^2+5 x-20&=-x^2+3 x+36\\
2 x^2+ x^2+5 x-3 x-20-36&= 0\\
3 x^2+2 x-56&=0.
\end{aligned}
\end{equation} Apply the quadratic formula: \begin{equation}
\begin{aligned}
x&=\frac{-2 \pm \sqrt{2^2-4 \cdot 3(-56)}}{2 \cdot 3}\\
&=\frac{-2 \pm 26}{2 \cdot 3}\\
&=\frac{-1 \pm 13}{ 3}\\
\therefore x &= \frac{-1+13}{ 3}= 4\\
x &= \frac{-1-13}{ 3}= -\frac{14}{3}\\
y&=-(4)^2+3 \cdot (4)+36 = 32\\
y&=-\left(-\frac{14}{3}\right)^2+3 \cdot \left(-\frac{14}{3}\right)+36 \\
& = -\frac{196}{9}-14+36\\
&= \frac{-196+22\cdot 9}{9} = \frac{2}{9}.
\end{aligned}
\end{equation} The solution set is $$(4,32), \quad \left(-\frac{14}{3}, \frac{2}{9}\right) .$$
