Example 10 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text {. }
$$

Find the minimum value of $b^{2}-4 a c$.
(2004, "TRULY ${ }^{\circledR}$ Xinli Cup" National Junior High School Mathematics Competition)
Let $y=a x^{2}+b x+c$.
From $a<0$, we know $\Delta=b^{2}-4 a c>0$.
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.

Since $x_{1} x_{2}=\frac{c}{a}<0$, let's assume $x_{1}<x_{2}$.

Then $x_{1}<0<x_{2}$, and the axis of symmetry is $x=-\frac{b}{2 a} \leqslant 0$.
Figure 4
The graph is shown in Figure 4.
Thus, $\left|x_{2}\right|=\left|\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\right|$
$$
=\frac{b-\sqrt{b^{2}-4 a c}}{2 a}=c \text {. }
$$

Therefore, $\frac{4 a c-b^{2}}{4 a} \geqslant c=\frac{b-\sqrt{b^{2}-4 a c}}{2 a} \geqslant-\frac{\sqrt{b^{2}-4 a c}}{2 a}$.
So, $b^{2}-4 a c \geqslant 4$.
When $a=-1, b=0, c=1$, the equality holds.
Thus, the minimum value of $b^{2}-4 a c$ is 4.