Find roots of polynomial $$ p(x) = x^2+99x+27 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -0.2735\\[1 em]x_2 &= -98.7265 \end{aligned} $$Explanation
The solutions of $ x^2+99x+27 = 0 $ are: $ x = -\dfrac{ 99 }{ 2 }-\dfrac{ 3 \sqrt{ 1077}}{ 2 } ~ \text{and} ~ x = -\dfrac{ 99 }{ 2 }+\dfrac{ 3 \sqrt{ 1077}}{ 2 }$.
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