Find roots of polynomial $$ p(x) = x^2+484x+39200 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -102.8454\\[1 em]x_2 &= -381.1546 \end{aligned} $$Explanation
The solutions of $ x^2+484x+39200 = 0 $ are: $ x = -242-2 \sqrt{ 4841 } ~ \text{and} ~ x = -242+2 \sqrt{ 4841 }$.
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