Question

\[17 x^2 + 28x + 12 = 0\]

Answer 1

Given: 

\[17 x^2 + 28x + 12 = 0\]

Comparing the given equation with the general form of the quadratic equation 

\[a x^2 + bx + c = 0\], we get

\[a = 17, b = 28\] and \[c = 12\]

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

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

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

\[\Rightarrow \alpha = \frac{- 28 + \sqrt{- 32}}{34}\] and \[\beta = \frac{- 28 - \sqrt{- 32}}{34}\]

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

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

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

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