Answer
$(1,-2), \quad \left(-\frac{31}{12}, -\frac{139}{48}\right) $
Work Step by Step
Given $$\begin{cases}
y=3 x^2+5 x-10 \\
y=\frac{1}{4} x-\frac{9}{4}.
\end{cases}$$ Set the two functions equal and solve for $x$.
\begin{equation}
\begin{aligned}
\left(3 x^2+5 x-10 \right)\cdot 4&=\left(\frac{1}{4} x-\frac{9}{4}\right)\cdot 4\\
12x^2+20x-40&= x-9\\
12x^2+19x-31&=0.
\end{aligned}
\end{equation} Apply the quadratic formula: \begin{equation}
\begin{aligned}
x&=\frac{-19 \pm \sqrt{19^2-4 \cdot 12(-31)}}{2 \cdot 12}\\
&=\frac{-19 \pm 43}{24}\\
\therefore x &= \frac{-19+43}{24}= 1\\
x&= \frac{-19-43}{24}= -\frac{31}{12}\\
y&=\frac{1}{4} \cdot 1-\frac{9}{4} = -2\\
y&=\frac{1}{4} \cdot \left( -\frac{31}{12} \right)-\frac{9}{4} = -\frac{139}{48}.
\end{aligned}
\end{equation} The solution set is $(1,-2), \quad \left(-\frac{31}{12}, -\frac{139}{48}\right) $
