Find roots of polynomial $$ p(x) = 3s^2+22s+10 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -0.4869\\[1 em]s_2 &= -6.8465 \end{aligned} $$Explanation
The solutions of $ 3s^2+22s+10 = 0 $ are: $ s = -\dfrac{ 11 }{ 3 }-\dfrac{\sqrt{ 91 }}{ 3 } ~ \text{and} ~ s = -\dfrac{ 11 }{ 3 }+\dfrac{\sqrt{ 91 }}{ 3 }$.
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