Solve the following

Given: $3 x^{2}-4 x+\frac{20}{3}=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x=c=0$, we get $a=3, b=-4$ and $c=\frac{20}{3}$.

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:

$\Rightarrow \alpha=\frac{4+\sqrt{16-4 \times 3 \times \frac{20}{3}}}{6}$ and $\beta=\frac{4-\sqrt{16-4 \times 3 \times \frac{20}{3}}}{6}$

$\Rightarrow \alpha=\frac{4+\sqrt{-64}}{6}$ and $\beta=\frac{4-\sqrt{-64}}{6}$

$\Rightarrow \alpha=\frac{4+8 i}{6} \quad$ and $\quad \beta=\frac{4-8 i}{6}$

$\Rightarrow \alpha=\frac{2+4 i}{3} \quad$ and $\quad \beta=\frac{2-4 i}{3}$

Hence, the roots of the equation are $\frac{2 \pm 4 i}{3}$.