Solve the following equation using quadratic formula:

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

Comparing equation $\sqrt{3}x^2 + 10x - 8\sqrt{3}$ = 0 with ax² + bx + c = 0, we get :

a = $\sqrt{3}$, b = 10 and c = $-8\sqrt{3}$.

By formula,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow x = \dfrac{-(10) \pm \sqrt{(10)^2 - 4 \times \sqrt{3} \times (-8\sqrt{3})}}{2\times(\sqrt{3})} \\[1em] = \dfrac{-10 \pm \sqrt{100 + 96}}{2\sqrt{3}} \\[1em] = \dfrac{-10 \pm \sqrt{196}}{2\sqrt{3}} \\[1em] = \dfrac{-10 \pm 14}{2\sqrt{3}} \\[1em] = \dfrac{2(-5 \pm 7)}{2\sqrt{3}} \\[1em] = \dfrac{-5 \pm 7}{\sqrt{3}} \\[1em] = \dfrac{-5 + 7}{\sqrt{3}} \text{ or } \dfrac{-5 - 7}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-12}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } \dfrac{-4 \times 3}{\sqrt{3}} \\[1em] = \dfrac{2}{\sqrt{3}} \text{ or } -4\sqrt{3}.$

Hence, $x = \Big{\dfrac{2}{\sqrt{3}}, -4\sqrt{3}\Big}$.