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

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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.
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