2. Given that $\tan \alpha, \tan \beta$ are the roots of the equation $x^{2}+3 \sqrt{3} x+4=0$, and $\alpha, \beta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \alpha+\beta$ is ( ).
A. $\frac{\pi}{3}$
B. $\frac{\pi}{3},-\frac{2 \pi}{3}$
C. $-\frac{\pi}{3}$ or $\frac{2 \pi}{3}$
D. $-\frac{2 \pi}{3}$
2. B. $\tan \alpha+\tan \beta=-3 \sqrt{3}, \tan \alpha \cdot \tan \beta=4, \tan (\alpha+\beta)=\frac{-3 \sqrt{3}}{1-4}=\sqrt{3}, \alpha+\beta \in(-\pi, \pi)$, so $\alpha+\beta=\frac{\pi}{3}$ or $\alpha+\beta=-\frac{2}{3} \pi$.