Question

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

Answer 1

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