Solve the following

Given: $17 x^{2}+28 x+12=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=17, b=28$ and $c=12$.

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:

$\alpha=\frac{-28+\sqrt{784-4 \times 17 \times 12}}{34}$ and $\beta=\frac{-28-\sqrt{784-4 \times 17 \times 12}}{34}$

$\Rightarrow \alpha=\frac{-28+\sqrt{784-816}}{34} \quad$ and $\quad \beta=\frac{-28-\sqrt{784-816}}{34}$

$\Rightarrow \alpha=\frac{-28+\sqrt{-32}}{34} \quad$ and $\quad \beta=\frac{-28-\sqrt{-32}}{34}$

$\Rightarrow \alpha=\frac{-28+\sqrt{32 i^{2}}}{34} \quad$ and $\quad \beta=\frac{-28-\sqrt{32 i^{2}}}{34}$

$\Rightarrow \alpha=\frac{-28+4 \sqrt{2} i}{34} \quad$ and $\quad \beta=\frac{-28-4 \sqrt{2} i}{34}$

$\Rightarrow \alpha=\frac{-14+2 \sqrt{2} i}{17} \quad$ and $\quad \beta=\frac{-14-2 \sqrt{2} i}{17}$

Hence, the roots of the equation are $-\frac{14}{17} \pm \frac{2 \sqrt{2}}{17} i$.