Example 5. If the parabola $y=ax^2+bx+c$ intersects the $x$-axis at $A$ and $B$, and the vertex is $C\left(-\frac{1}{4},-6 \frac{1}{8}\right)$, and the area of $\triangle ABC$ is $10 \frac{23}{32}$. Try to find the equation of the parabola.
Solve for the vertex coordinates of $y=a x^{2}+b x+c$ as
$\mathrm{C}\left(-\frac{\mathrm{b}}{2 \mathrm{a}}, \frac{4 \mathrm{ac}-\mathrm{b}^{2}}{4 \mathrm{a}}\right)$,
thus $-\frac{\mathrm{b}}{2 \mathrm{a}}=-\frac{1}{4}$,
$$
\begin{array}{c}
\frac{4 \mathrm{ac}-\mathrm{b}^{2}}{4 \mathrm{a}}=-6 \frac{1}{8} . \\
\because \mathrm{S}_{\triangle \mathrm{ABC}}=\frac{1}{2}|\mathrm{AB}| \cdot \mathrm{h}_{0}=\frac{1}{2}|\mathrm{AB}| \\
\cdot 6 \frac{1}{8}=10 \frac{23}{32} . \\
\therefore \quad|\mathrm{AB}|=\frac{7}{2} . \\
\frac{\sqrt{\mathrm{b}^{2}-4 \mathrm{ac}}}{|\mathrm{a}|}=\frac{7}{2} .
\end{array}
$$

Solving (1), (2), and (3) simultaneously, we get
$$
\mathrm{a}=2, \mathrm{~b}=1, \mathrm{c}=-6 \text {. }
$$

Therefore, the equation of the parabola is $y=2 x^{2}+x-6$.