Question

Solve $21x^2-28x+10=0$

Answer 1

Here $21x^2-28x+10=0$

Comparing the given quadratic equation with $a{x}^2+bx+c=0$ we have

$a=21,b=-28$ and $c=10$

$\therefore x=\frac{-(-28)\pm \sqrt{(-28{)}^2-4\times 21\times 10}}{2\times 21}$

$=\frac{28\pm \sqrt{784-840}}{42}=\frac{28\pm \sqrt{-56}}{42}$

$=\frac{28\pm 2\sqrt{14}i}{42}=\frac{14\pm 2\sqrt{14}i}{21}$

Thus $x=\frac{2}{3}+\frac{\sqrt{14}}{21}iandx=\frac{2}{3}-\frac{\sqrt{14}}{21}i$.