Answer
No real solution.
Work Step by Step
Given $$
\begin{cases}
& y=x^2+2 x-8 \\
& y=-2 x^2+2 x-18.
\end{cases}
$$ Set the two equations equal and solve for the values of $x$.
$$
\begin{aligned}
x^2+2 x-8 & =-2 x^2+2 x-18 \\
x^2+2 x^2+2 x-2 x & =8-18 \\
3 x^2 & =-10 \\
x^2 & =\frac{-10}{3}.
\end{aligned}
$$ Since $x^2$ must be positive, it follows that the given system has no real solution.
