Question:
$3 x^{2}-4 x+\frac{20}{3}=0$
Solution:
Given:
$3 x^{2}-4 x+\frac{20}{3}=0$
Multiplying both the sides by 3 we get,
$9 x^{2}-12 x+20=0$
Solution of a general quadratic equation $a x^{2}+b x+c=0$ is given by:
$x=\frac{-b \pm \sqrt{\left(b^{2}-4 a c\right)}}{2 a}$
$\Rightarrow x=\frac{-(-12) \pm \sqrt{(-12)^{2}-(4 \times 9 \times 20)}}{2 \times 9}$
$\Rightarrow x=\frac{12 \pm \sqrt{144-720}}{18}$
$\Rightarrow x=\frac{12 \pm \sqrt{-576}}{18}$
$\Rightarrow x=\frac{12 \pm 24 i}{18}$
$x=\frac{12}{18} \pm \frac{24}{18} i$
$\Rightarrow \quad x=\frac{2}{3} \pm \frac{4}{3} i$
Ans: $x=\frac{2}{3}+\frac{4}{3} i$ and $x=\frac{2}{3}-\frac{4}{3} i$