Given:

$\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$

On comparing it with $a x^{2}+b x+x=0$, we get; $a=\sqrt{3}, b=10$ and $c=-8 \sqrt{3}$

Discriminant $D$ is given by:

$\begin{array}{l}
D=\left(b^{2}-4 a c\right) \\
=(10)^{2}-4 \times \sqrt{3} \times(-8 \sqrt{3}) \\
=100+96 \\
=196>0
\end{array}$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by:

$\begin{array}{l}
\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-10+\sqrt{196}}{2 \sqrt{3}}=\frac{-10+14}{2 \sqrt{3}}=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(10)-\sqrt{196}}{2 \sqrt{3}}=\frac{-10-14}{2 \sqrt{3}}=\frac{-24}{2 \sqrt{3}}=\frac{-12}{\sqrt{3}}=\frac{-12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-12 \sqrt{3}}{3}=-4 \sqrt{3}
\end{array}$

Thus, the roots of the equation are $\frac{2 \sqrt{3}}{3}$ and $-4 \sqrt{3}$.