Solve the equation


Solve the equation $3 x^{2}-4 x+\frac{20}{3}=0$


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

$\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}=\frac{-(-12) \pm \sqrt{-576}}{2 \times 9}=\frac{12 \pm \sqrt{576} i}{18}$   $[\sqrt{-1}=i]$

$=\frac{12 \pm 24 i}{18}=\frac{6(2 \pm 4 i)}{18}=\frac{2 \pm 4 i}{3}=\frac{2}{3} \pm \frac{4}{3} i$

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