2. Let $a$, $b$, $c$, and $S$ be the lengths of the three sides and the area of a triangle, respectively. For the equation in $x$
$$
b^{2} x^{2}+\left(b^{2}+c^{2}-a^{2}\right) x+c^{2}=0
$$

the discriminant is $\Delta$. Then the relationship between $\Delta$ and $S$ is ( ).
(A) $\Delta=16 S^{2}$
(B) $\Delta=-16 S^{2}$
(C) $\Delta=16 S$
(D) $\Delta=-16 S$
2.B.

Since $\Delta=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}$
$$
\begin{array}{l}
=\left(b^{2}+c^{2}-a^{2}+2 b c\right)\left(b^{2}+c^{2}-a^{2}-2 b c\right) \\
=\left[(b+c)^{2}-a^{2}\right]\left[(b-c)^{2}-a^{2}\right] \\
=(b+c+a)(b+c-a)(b-c+a)(b-c-a) .
\end{array}
$$

Let $p=\frac{1}{2}(a+b+c)$, so,
$$
\begin{array}{l}
\Delta=2 p \cdot 2(p-a) \cdot 2(p-c)[-2(p-b)] \\
=-16 p(p-a)(p-b)(p-c) .
\end{array}
$$

By Heron's formula, $S^{2}=p(p-a)(p-b)(p-c)$. Therefore, $\Delta=-16 S^{2}$.