$$
y=a x^{2}+b x+c(a \neq 0)
$$

The image vertex is $C$, and the two intersections with the $x$-axis are $A$ and $B$. If $\triangle A B C$ is an acute triangle, and $\sin \angle B C A$ $=\frac{4}{5}$, then $\Delta=b^{2}-4 a c=$ . $\qquad$
2. 16 .

From $A C=B C$ and $\sin \angle B C A=\frac{4}{5}$, we know $\tan B=2$.
Draw $C H \perp A B$ at point $H$.
Then $C H=\left|y_{C}\right|=\frac{\Delta}{4 a}, B H=\frac{A B}{2}=\frac{\sqrt{\Delta}}{2 a}$.

Thus $\tan B=\frac{C H}{B H}=\frac{\sqrt{\Delta}}{2}=2$
$$
\Rightarrow \Delta=16 \text {. }
$$