The given equation is $2 x^{2}+6 \sqrt{3} x-60=0$.

Comparing it with $a x^{2}+b x+c=0$, we get $a=2, b=6 \sqrt{3}$ and $c=-60$

$\therefore$ Discriminant,
$D=b^{2}-4 a c=(6 \sqrt{3})^{2}-4 \times 2 \times(-60)=180+480=588>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{588}=14 \sqrt{3}$

$\begin{array}{l}
\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-6 \sqrt{3}+14 \sqrt{3}}{2 \times 2}=\frac{8 \sqrt{3}}{4}=2 \sqrt{3} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-6 \sqrt{3}-14 \sqrt{3}}{2 \times 2}=\frac{-20 \sqrt{3}}{4}=-5 \sqrt{3}
\end{array}$

Hence, $2 \sqrt{3}$ and $-5 \sqrt{3}$ are the root of the given equation.