Answer
$(2(1+\sqrt 3),5(1+2\sqrt 3));(2(1-\sqrt 3),5(1-2\sqrt 3))$
Work Step by Step
Given \begin{equation}
\begin{aligned}
& y=0.25 x^2+4 x-7 \\
& y=0.5 x^2+3 x-9
\end{aligned}
\end{equation} Set the two equations equal and find the values of $x$. Use the values of $x$ to find the corresponding values of $y$ and write the solutions as points, $(x,y)$.
\begin{equation}
\begin{aligned}
0.5 x^2+3 x-9&=0.25 x^2+4 x-7 \\
(0.5-0.25) x^2+(3-4) x+(-9+7)&=0\\
\frac{0.25 x^2-x-2}{0.25}&=\frac{0}{0.25}\\
x^2-4 x-8&=0.
\end{aligned}
\end{equation} Use the method of completing the square: \begin{equation}
\begin{aligned}
x^2-4 x+2^2 & =8+2^2 \\
(x-2)^2 & =12 \\
(x-2)^2 & = \pm \sqrt{4 \cdot 3} \\
(x-2)^2 & = \pm 2 \sqrt{3}\\
x&=2 \pm 2 \sqrt{3}.
\end{aligned}
\end{equation} This gives \begin{equation}
\begin{aligned}
& x=2+2 \sqrt{3}=2(1+\sqrt 3) \\
& x=2-2 \sqrt{3}=2(1-\sqrt 3).
\end{aligned}
\end{equation} The corresponding $y$ values are:
\begin{equation}
\begin{aligned}
& y=0.25(2+2\sqrt 3)^2+4(2+2\sqrt 3)-7=5+10\sqrt 3=5(1+2\sqrt 3) \\
& y=0.25(2-2\sqrt 3)^2+4(2-2\sqrt 3)-7=5-10\sqrt 3=5(1-2\sqrt 3).
\end{aligned}
\end{equation} Solution: $$(2(1+\sqrt 3),5(1+2\sqrt 3));(2(1-\sqrt 3),5(1-2\sqrt 3)).$$
