Answer
No real solution
Work Step by Step
Given $$
\begin{cases}
& y=-0.5 x^2+4 x-8 \\
& y=0.3 x^2+5 x+7.
\end{cases}
$$ Set the two equations equal and solve for the values of $x$. $$
\begin{aligned}
-0.5 x^2+4 x-8&=0.3 x^2+5 x+7\\
-0.5 x^2-0.3 x^2+4 x-5 x & =7+8 \\
-0.8 x^2-x & =15 \\
(-10)\left(-0.8 x^2-x-15\right) & =0 \\
8 x^2+10 x+150 & =0
\end{aligned}
$$ $$
\begin{aligned}
& a=8 \\
& b=-10 \\
& c=150
\end{aligned}
$$ $$
\begin{aligned}
x& =\frac{-10 \pm \sqrt{(10)^2-4(8)(150)}}{2(8)}\\
x& =\frac{-10 \pm \sqrt{-4700} }{16}.
\end{aligned}
$$ As there is no real solution for $x$, it means the given system has no real solution.
