Answer
No real solution
Work Step by Step
Given $$
\begin{cases}
& y=x^2-10 x+30 \\
& y=-x^2+6 x-15.\end{cases}
$$ Set the two equations equal and solve for the values of $x$.
$$
\begin{aligned}
x^2-10 x+30 & =-x^2+6 x-15 \\
x^2+x^2-10 x-6 x & =-15-30 \\
2 x^2-16 x & =-45 \\
2 x^2-16 x+45 & =0
\end{aligned}
$$ $$
\begin{aligned}
& a=2 \\
& b=-16 \\
& c=45
\end{aligned}
$$ $$
\begin{aligned}
x& =\frac{-(-16) \pm \sqrt{(-16)^2-4 \cdot 2 \cdot 45}}{2 \cdot 2}\\
x& =\frac{16 \pm \sqrt{-104} }{4}.
\end{aligned}
$$ Since the equation has no real solution, it follows that the given system has no real solution.
