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

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Hint: Compare the equation with the general quadratic equation, then use the formula for finding out the roots of a quadratic equation.

The given quadratic equation is $3{x^2} - 4x + \frac{{20}}{3} = 0$,

Comparing it with general quadratic equation, $a{x^2} + bx + c = 0$, we have:

$a = 3,b = - 4$ and $c = \frac{{20}}{3}$

And we know that the roots of quadratic equation is given as:

$\alpha ,\beta = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substituting the values from the above equation, we have:

$

\Rightarrow \alpha ,\beta = \frac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 3 \times \frac{{20}}{3}} }}{{2(3)}}, \\

\Rightarrow \alpha ,\beta = \frac{{4 \pm \sqrt {16 - 80} }}{6} = \frac{{4 \pm \sqrt { - 64} }}{6}, \\

\Rightarrow \alpha ,\beta = \frac{{4 \pm 8i}}{6} = \frac{{2 \pm 4i}}{3} \\

$

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

Thus the roots of the equation are $\frac{2}{3} + \frac{4}{3}i$ and $\frac{2}{3} - \frac{4}{3}i$

Note: Discriminant of a quadratic equation is:

$ \Rightarrow D = {b^2} - 4ac$

If the discriminant of a quadratic equation is less than zero (i.e. negative), the roots of the equation will always be imaginary and they will be complex conjugates of each other.

The given quadratic equation is $3{x^2} - 4x + \frac{{20}}{3} = 0$,

Comparing it with general quadratic equation, $a{x^2} + bx + c = 0$, we have:

$a = 3,b = - 4$ and $c = \frac{{20}}{3}$

And we know that the roots of quadratic equation is given as:

$\alpha ,\beta = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Substituting the values from the above equation, we have:

$

\Rightarrow \alpha ,\beta = \frac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 3 \times \frac{{20}}{3}} }}{{2(3)}}, \\

\Rightarrow \alpha ,\beta = \frac{{4 \pm \sqrt {16 - 80} }}{6} = \frac{{4 \pm \sqrt { - 64} }}{6}, \\

\Rightarrow \alpha ,\beta = \frac{{4 \pm 8i}}{6} = \frac{{2 \pm 4i}}{3} \\

$

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

Thus the roots of the equation are $\frac{2}{3} + \frac{4}{3}i$ and $\frac{2}{3} - \frac{4}{3}i$

Note: Discriminant of a quadratic equation is:

$ \Rightarrow D = {b^2} - 4ac$

If the discriminant of a quadratic equation is less than zero (i.e. negative), the roots of the equation will always be imaginary and they will be complex conjugates of each other.

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