2. For the quadratic function $y=a x^{2}+b x+c$, the graph intersects the $x$-axis at two distinct points $A$ and $B$, and the vertex of the parabola is $C$. Then $S_{\triangle A B C}=$ $\qquad$
2. $\frac{\left(b^{2}-4 a c\right) \sqrt{b^{2}-4 a c}}{8 a^{2}}$.

Since $\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{|a|}$, the coordinates of the vertex are $C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$, therefore,
$$
\begin{aligned}
& S_{\triangle B C}=\frac{1}{2} \cdot \frac{\sqrt{b^{2}-4 a c}}{|a|} \cdot\left|\frac{4 a c-b^{2}}{4 a}\right| \\
= & \frac{\left(b^{2}-4 a c\right) \sqrt{b^{2}-4 a c}}{8 a^{2}} .
\end{aligned}
$$