Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is $2 x^{2}+5 \sqrt{3} x+6=0$.

Comparing it with $a x^{2}+b x+c=0$, we get

$a=2, b=5 \sqrt{3}$ and $c=6$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(5 \sqrt{3})^{2}-4 \times 2 \times 6=75-48=27>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{27}=3 \sqrt{3}$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-5 \sqrt{3}+3 \sqrt{3}}{2 \times 2}=\frac{-2 \sqrt{3}}{4}=-\frac{\sqrt{3}}{2}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-5 \sqrt{3}-3 \sqrt{3}}{2 \times 2}=\frac{-8 \sqrt{3}}{4}=-2 \sqrt{3}$

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