Answer
$\left(-\frac{5}{3},-\frac{143}{9}\right)$ and $(3.5,68.5)$
Work Step by Step
Given $$
\begin{cases}
& y=4 x^2+9 x-12 \\
& y=-2 x^2+20 x+23.
\end{cases}
$$ Set the two equations equal and solve for the values of $x$.
$$
\begin{aligned}
& 4 x^2+9 x-12=-2 x^2+20 x+23 \\
& 4 x^2+2 x^2+9 x-20 x=23+12 \\
& 6 x^2-11 x=35 \\
& 6 x^2-11 x-35=0
\end{aligned}
$$ $$
\begin{array}{r}
a=6 \\
b=-11\\
c=-35
\end{array}
$$ $$
\begin{aligned}
& x=\frac{-(-11) \pm \sqrt{(-11)^2-4 \cdot 6 \cdot (-35)}}{2 \cdot 6} \\
& x=\frac{11 \pm \sqrt{961}}{12} \\
& x=\frac{11 \pm 31}{12} \\
\end{aligned}
$$ $$
\begin{aligned}
x & =\frac{11 - 31}{12} \\
& =-\frac{5}{3} \\
&\approx -1.66667\\
x & =\frac{11 + 31}{12} \\
& =3.5.
\end{aligned}
$$ Find the corresponding $y$ values using either of the given equations.
$$
\begin{aligned}
y&= 4\left(-\frac{5}{3}\right)^2+9\cdot\left(-\frac{5}{3}\right) -12\\
&=-\frac{143}{9}\\
& \approx -15.888889\\
y&= 4(3.5)^2+9\cdot( 3.5) -12\\
& =68.5.
\end{aligned}
$$ Plot the two functions in the same window to check the solution(s).
The solution is $\left(-\frac{5}{3},-\frac{143}{9}\right)$ and $(3.5,68.5)$..
