Question:
Solve the equation $3 x^{2}-4 x+\frac{20}{3}=0$
Solution:
The given quadratic equation is $3 x^{2}-4 x+\frac{20}{3}=0$
This equation can also be written as $9 x^{2}-12 x+20=0$
On comparing this equation with $a x^{2}+b x+c=0$, we obtain
$a=9, b=-12$, and $c=20$
Therefore, the discriminant of the given equation is
$D=b^{2}-4 a c=(-12)^{2}-4 \times 9 \times 20=144-720=-576$
Therefore, the required solutions are
$\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-(-12) \pm \sqrt{-576}}{2 \times 9}=\frac{12 \pm \sqrt{576} i}{18}$ $[\sqrt{-1}=i]$
$=\frac{12 \pm 24 i}{18}=\frac{6(2 \pm 4 i)}{18}=\frac{2 \pm 4 i}{3}=\frac{2}{3} \pm \frac{4}{3} i$