Find roots of polynomial $$ p(x) = s^2+3s+25 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -\frac{ 3 }{ 2 }+\frac{\sqrt{ 91 }}{ 2 }i\\[1 em]s_2 &= -\frac{ 3 }{ 2 }- \frac{\sqrt{ 91 }}{ 2 }i \end{aligned} $$Explanation
The solutions of $ s^2+3s+25 = 0 $ are: $ s = -\dfrac{ 3 }{ 2 }+\dfrac{\sqrt{ 91 }}{ 2 }i ~ \text{and} ~ s = -\dfrac{ 3 }{ 2 }-\dfrac{\sqrt{ 91 }}{ 2 }i$.
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