2. If $a \geqslant b \geqslant c, a+b+c=0$, and $x_{1}, x_{2}$ are the two real roots of the quadratic equation $a x^{2}+b x+c=0$, then the sum of the maximum and minimum values of $\left|x_{1}^{2}-x_{2}^{2}\right|$ is $\qquad$.
$$
\begin{array}{l}
\left|x_{1}^{2}-x_{2}^{2}\right|=\left|\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)\right|=\left|\frac{b}{a} \cdot \frac{\sqrt{b^{2}-4 a c}}{a}\right|=\left|\frac{b \sqrt{b^{2}+4 a(a+b)}}{a^{2}}\right| \\
=\left|\frac{b(2 a+b)}{a^{2}}\right|=\left|\left(\frac{b}{a}\right)^{2}+2 \cdot \frac{b}{a}\right|, \text { by }\left\{\begin{array}{l}
a \geqslant b, \\
b \geqslant-(a+b), \Rightarrow-\frac{1}{2} \leqslant \frac{b}{a} \leqslant 1, \\
a>0
\end{array}\right.
\end{array}
$$

Therefore, $\left(\frac{b}{a}\right)^{2}+2 \cdot \frac{b}{a} \in\left[-\frac{3}{4}, 3\right] \Rightarrow\left|x_{1}^{2}-x_{2}^{2}\right| \in[0,3]$, the sum of its maximum and minimum values is 3.