Question

Solve the equation ${21x}^2-28x+10=0$

Answer 1

The given quadratic equation is, ${21x}^2-28x+10=0$
On comparing this equation with ${ax}^2+bx+c=0$,
we obtain $a=21,b=-28,$ and $c=10$
Therefore, the discriminant of the given equation is
$D=b^2-4ac={(-28)}^2-4\times 21\times 10=784-840=-56$
Therefore, the required solutions are
$\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-28)\pm \sqrt{-56}}{2\times 21}=\frac{28\pm \sqrt{56}i}{42}$
$=\frac{28\pm 2\sqrt{14}i}{42}=\frac{28}{42}\pm \frac{2\sqrt{14}}{42}i=\frac{2}{3}\pm \frac{\sqrt{14}}{21}i$