Question
Find roots of polynomial $$ p(x) = 17y^2+2y+6 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= -\dfrac{ 1 }{ 17 }+\dfrac{\sqrt{ 101 }}{ 17 }i\\[1 em]y_2 &= -\dfrac{ 1 }{ 17 }- \dfrac{\sqrt{ 101 }}{ 17 }i \end{aligned} $$Explanation
The solutions of $ 17y^2+2y+6 = 0 $ are: $ y = -\dfrac{ 1 }{ 17 }+\dfrac{\sqrt{ 101 }}{ 17 }i ~ \text{and} ~ y = -\dfrac{ 1 }{ 17 }-\dfrac{\sqrt{ 101 }}{ 17 }i$.
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