Find roots of polynomial $$ p(x) = 100s^2+72s+144 $$
Answer
The roots of polynomial $ p(s) $ are:
$$ \begin{aligned}s_1 &= -\frac{ 9 }{ 25 }+\frac{ 3 \sqrt{ 91}}{ 25 }i\\[1 em]s_2 &= -\frac{ 9 }{ 25 }-3 \frac{\sqrt{ 91 }}{ 25 }i \end{aligned} $$Explanation
The solutions of $ 100s^2+72s+144 = 0 $ are: $ s = -\dfrac{ 9 }{ 25 }+\dfrac{ 3 \sqrt{ 91}}{ 25 }i ~ \text{and} ~ s = -\dfrac{ 9 }{ 25 }-\dfrac{ 3 \sqrt{ 91}}{ 25 }i$.
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