Solve the following


$13 x^{2}+7 x+1=0$


Given: $13 x^{2}+7 x+1=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=13, b=7$ 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:

$\alpha=\frac{-7+\sqrt{49-4 \times 13 \times 1}}{2 \times 13}$ and $\beta=\frac{-7-\sqrt{49-4 \times 13 \times 1}}{2 \times 13}$

$\Rightarrow \alpha=\frac{-7+\sqrt{49-52}}{26} \quad$ and $\quad \beta=\frac{-7-\sqrt{49-52}}{26}$

$\Rightarrow \alpha=\frac{-7+\sqrt{-3}}{26} \quad$ and $\quad \beta=\frac{-7-\sqrt{-3}}{26}$

$\Rightarrow \alpha=\frac{-7+i \sqrt{3}}{26} \quad$ and $\quad \beta=\frac{-7-i \sqrt{3}}{26}$

$\Rightarrow \alpha=-\frac{7}{26}+\frac{\sqrt{3}}{26} i \quad$ and $\quad \beta=-\frac{7}{26}-\frac{\sqrt{3}}{26} i$

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