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

Find the minimum value of $b^{2}-4 a c$.
Solution 1: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2 a c}=-1(a c \neq 0),
$$

This indicates that the quadratic equation
$$
a c x^{2}+b x+1=0
$$

has a real root $x=-1$. Substituting into equation (1) yields
$$
a c=b-1 \text {. }
$$

Transform equation (3) into the overall structure of the discriminant
$$
b^{2}-4 a c=(2 a c x+b)^{2},
$$

Substitute $x=-1, a c=b-1$, and note that $b \leqslant 0$, we get
$$
\begin{array}{l}
b^{2}-4 a c=[2(b-1)(-1)+b]^{2} \\
=(2-b)^{2} \geqslant 4 .
\end{array}
$$

When $b=0$, thus $a c=-1$, $b^{2}-4 a c$ reaches its minimum value of 4.

Note 1: In the above simplified proof, equation (4) plays a crucial role, indicating that the explicit representation of the essential relationship is a source of "ingenious and clever solutions." The entire solution can be completed mentally.

Note 2: The above solution does not use the conditions $a<0, c<0$, but only uses $a c \neq 0$, so the conditions can be weakened.

Solution 2: The given condition is
$$
\frac{-b+\sqrt{b^{2}-4 a c}}{2}=-a c \text {. }
$$

This indicates that the quadratic equation
$$
x^{2}+b x+a c=0
$$

has a real root $x_{1}=-a c$. Therefore,
$$
x_{2}=\frac{a c}{x_{1}}=-1 \text { (using } a c \neq 0 \text { ). }
$$

Substituting $x_{2}=-1$ into equation @ yields
$$
1-b+a c=0 \text {. }
$$

At this point, by $b \leqslant 0$ we know
$$
a c=b-1
$$
$\leqslant-1$ (ensured by $b \leqslant 0$ that $a c<0$ ),
so,
$$
1-a c \geqslant 2 \text {. }
$$

Substituting equations (1) and (9) into the discriminant, we have
$$
\begin{array}{l}
b^{2}-4 a c=(1+a c)^{2}-4 a c \\
=(1-a c)^{2} \geqslant 2^{2} .
\end{array}
$$

From equation (9), we know that when $a c=-1, b=0$, $b^{2}-4 a c$ can reach its minimum value of 4.