Answer
$(-5,25),(3,9)$
Work Step by Step
Set $y=x^2$ equal to $y=15-2 x$ and solve
$$
\begin{aligned}
x^2 & =15-2 x \\
x^2+2 x-15 & =0 \\
(x+5)(x-3) & =0 \\
x & =-5,x=3
\end{aligned}
$$
$x=-5 \quad$ and $\quad y=15-2(-5)=25 \quad$ or $x=3 \quad$ and $\quad y=15-2(3)=9$
The points of intersection are $(-5,25),(3,9)$.
