Solve the following


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


Given: $21 x^{2}-28 x+10=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=21, b=-28$ and $c=10$.

Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:

$\alpha=\frac{28+\sqrt{784-4 \times 21 \times 10}}{2 \times 21}$ and $\beta=\frac{28-\sqrt{784-4 \times 21 \times 10}}{2 \times 21}$

$\Rightarrow \alpha=\frac{28+\sqrt{-56}}{42} \quad$ and $\quad \beta=\frac{28-\sqrt{-56}}{42}$

$\Rightarrow \alpha=\frac{28+2 i \sqrt{14}}{42}$ and $\beta=\frac{28-2 i \sqrt{14}}{42}$

$\Rightarrow \alpha=\frac{14+i \sqrt{14}}{21}$ and $\beta=\frac{14-i \sqrt{14}}{21}$

$\Rightarrow \alpha=\frac{2}{3}+\frac{\sqrt{14}}{21} i$ and $\beta=\frac{2}{3}-\frac{\sqrt{14}}{21} i$

Hence, the roots of the equation are $\frac{2}{3} \pm \frac{\sqrt{14}}{21}$.

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