$21 x^{2}+9 x+1=0$
Given: $21 x^{2}+9 x+1=0$
Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=21, b=9$ and $c=1$.
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:
$\alpha=\frac{-9+\sqrt{81-4 \times 21 \times 1}}{2 \times 21} \quad$ and $\quad \beta=\frac{-9-\sqrt{81-4 \times 21 \times 1}}{2 \times 21}$
$\Rightarrow \alpha=\frac{-9+\sqrt{3} i}{42} \quad$ and $\quad \beta=\frac{-9-\sqrt{3} i}{42}$
$\Rightarrow \alpha=-\frac{9}{42}+\frac{\sqrt{3} i}{42} \quad$ and $\quad \beta=-\frac{9}{42}-\frac{\sqrt{3} i}{42}$
$\Rightarrow \alpha=-\frac{3}{14}+\frac{\sqrt{3} i}{42} \quad$ and $\quad \beta=-\frac{3}{14}-\frac{\sqrt{3} i}{42}$
Hence, the roots of the equation are $-\frac{3}{14} \pm \frac{i \sqrt{3}}{42}$.