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

Hint: We use the formula for the roots of quadratic equations. The formula for it is given by$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Here, $x$ is the root.

Complete step-by-step answer:
Now, consider the equation $21{{x}^{2}}-28x+10=0$
If we compare it with the standard form of the quadratic equation $a{{x}^{2}}+bx+c=0$ then it clearly implies that $a=21$ , $b=-28$ and $c=10$
Now, we use the formula for roots $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and substitute the values $a=21,$ $b=-28$ and $c=10$ in it.
 After this we get an equation like this,
$\begin{align}
  & x=\dfrac{-\left( -28 \right)\pm \sqrt{{{\left( -28 \right)}^{2}}-4\left( 21 \right)\left( 10 \right)}}{2\left( 21 \right)} \\
 & x=\dfrac{28\pm \sqrt{784-840}}{42} \\
 & x=\dfrac{28\pm \sqrt{-56}}{42} \\
\end{align}$
To open the negative root we will now use $\sqrt{-1}=i$ type form. This results in $\dfrac{28\pm \sqrt{-56}}{42}=\dfrac{28\pm 2i\sqrt{14}}{42}$ form of the above equation.
The $i$ in the above equation is called iota. The roots containing iota make a root to be known as a complex root instead of real root. Now, we take out the common value from numerator and denominator in order to get the required roots. The common factor from numerator and denominator is 2 here. So it implies that $\dfrac{28\pm 2i\sqrt{14}}{42}=\dfrac{2}{2}\left( \dfrac{14\pm i\sqrt{14}}{21} \right)$
It results into $\dfrac{14\pm i\sqrt{14}}{21}$
This implies $\dfrac{28\pm 2i\sqrt{14}}{42}=\dfrac{14\pm i\sqrt{14}}{21}$
So after solving the equation for x we get its two roots these are $\dfrac{14+i\sqrt{14}}{21}$ and $\dfrac{14-i\sqrt{14}}{21}$
Clearly, we can see that these roots are in complex form.
In a complex form the roots are negative inside the root. This makes the difference in real and complex roots.
Hence, the roots of the quadratic equation is given by $\dfrac{14\pm i\sqrt{14}}{21}$

Note: It is clear from the equation that we should use quadratic root formula and one has to keep in mind that the formula implies two roots by being quadratic. Do not stop after getting iota in question. Iota arises due to the negative sign inside the root. Keep on solving it and it will result in the right answer.