Find roots of polynomial $$ p(x) = 3s^2+2s+6 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -\frac{ 1 }{ 3 }+\frac{\sqrt{ 17 }}{ 3 }i\\[1 em]s_2 &= -\frac{ 1 }{ 3 }- \frac{\sqrt{ 17 }}{ 3 }i \end{aligned} $$Explanation
The solutions of $ 3s^2+2s+6 = 0 $ are: $ s = -\dfrac{ 1 }{ 3 }+\dfrac{\sqrt{ 17 }}{ 3 }i ~ \text{and} ~ s = -\dfrac{ 1 }{ 3 }-\dfrac{\sqrt{ 17 }}{ 3 }i$.
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