3. The graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $A$ and $B$, with the vertex at $C$. If $\triangle A C B$ is a right triangle, then the value of the discriminant is $\qquad$.
3. 4 .

As shown in Figure 4.
From the problem, we know $\Delta=b^{2}-4 a c>0$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$.
Then $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$
$=\left(-\frac{b}{a}\right)^{2}-\frac{4 c}{a}=\frac{b^{2}-4 a c}{a^{2}}$.
Assume $a>0$. Then
$A B=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{a}$.
Thus $\frac{\left|4 a c-b^{2}\right|}{4 a}=\frac{\sqrt{b^{2}-4 a c}}{2 a} \Rightarrow \sqrt{b^{2}-4 a c}=2$.
Therefore, $\Delta=b^{2}-4 a c=4$.