Question
Find roots of polynomial $$ p(x) = x^2+110x-200 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 1.7891\\[1 em]x_2 &= -111.7891 \end{aligned} $$Explanation
The solutions of $ x^2+110x-200 = 0 $ are: $ x = -55-5 \sqrt{ 129 } ~ \text{and} ~ x = -55+5 \sqrt{ 129 }$.
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