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