Question
Find roots of polynomial $$ p(x) = x^2+160x-2304 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 13.2952\\[1 em]x_2 &= -173.2952 \end{aligned} $$Explanation
The solutions of $ x^2+160x-2304 = 0 $ are: $ x = -80-16 \sqrt{ 34 } ~ \text{and} ~ x = -80+16 \sqrt{ 34 }$.
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