Question
Find roots of polynomial $$ p(x) = x^2+8x-1200 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 30.8712\\[1 em]x_2 &= -38.8712 \end{aligned} $$Explanation
The solutions of $ x^2+8x-1200 = 0 $ are: $ x = -4-8 \sqrt{ 19 } ~ \text{and} ~ x = -4+8 \sqrt{ 19 }$.
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