II. Try to derive the quadratic formula for the equation $a x^{2}+b x+c=0(a \neq 0)$.

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II. Solve $\quad \because a x^{2}+b x+c=0 \quad(a \neq 0)$.
$\therefore$ Both sides of the equation can be divided by $a$, yielding
$$
x^{2}+\frac{b}{a} x+\frac{c}{a}=0 \text {. }
$$

Rearranging, we get $x^{2}+\frac{b}{a} x=-\frac{c}{a}$.
Adding the square of half the coefficient of the linear term to both sides, we get
$$
x^{2}+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^{2}=-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2} \text {. }
$$

This simplifies to $\left(x+\frac{b}{2 a}\right)^{2}=\frac{b^{2}-4 a c}{4 a^{2}}$.
When $b^{2}-4 a c \geqslant 0$, we have
$$
\begin{aligned}
& x+\frac{b}{2 a}= \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}, \\
\therefore \quad & x=-\frac{b}{2 a} \pm \frac{\sqrt{b^{2}-4 a c}}{2 a}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .
\end{aligned}
$$