Example 3 Let $f(x)$ be a quadratic function, and $f(0)=12$,
$$
\begin{array}{c}
g(x)=2^{x} f(x), \\
g(x+1)-g(x) \geqslant 2^{x+1} x^{2} .
\end{array}
$$

Find the expressions for $f(x)$ and $g(x)$.
Given that $f(x)$ is a quadratic function, we can assume it to be $f(x)=a x^{2}+b x+12(a \neq 0)$ using the method of undetermined coefficients.
From (1) substituting into (2), we have $2^{x+1} f(x+1)-2^{x} f(x) \geqslant 2^{x+1} x^{2}$,
which simplifies to
$$
2 f(x+1)-f(x) \geqslant 2 x^{2}.
$$

Substituting (3) into (4), we get $a x^{2}+(4 a+b) x+(2 a+2 b+12) \geqslant 2 x^{2}$,
leading to (1)
Thus,
$$
\begin{array}{c}
a=2, b=-8. \\
f(x)=2 x^{2}-8 x+12, \\
g(x)=\left(2 x^{2}-8 x+12\right) 2^{x}.
\end{array}
$$

Verification confirms this is the solution.
Alternatively, if (2) $a \neq 2$, then $\left\{\begin{array}{l}a>2, \\ \Delta=(4 a+b)^{2}-4(a-2)(2 a+2 b+12) \leqslant 0 \text {. }\end{array}\right.$
This simplifies to $8(a+1)^{2}+(b-16)^{2} \leqslant 168(a>2), 2<a \leqslant \sqrt{21-\frac{1}{8}(b-16)^{2}}-1$.
In summary, $f(x)=2 x^{2}-8 x+12, g(x)=\left(2 x^{2}-8 x+12\right) 2^{x}$, or
$f(x)=a x^{2}+b x+12, g(x)=\left(a x^{2}+b x+12\right) 2^{x}$, where $a, b$ satisfy $2<a \leqslant$ $\sqrt{21-\frac{1}{8}(b-16)^{2}}-1$.