Find roots of polynomial $$ p(x) = x^2+100x-4800 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 35.44\\[1 em]x_2 &= -135.44 \end{aligned} $$Explanation
The solutions of $ x^2+100x-4800 = 0 $ are: $ x = -50-10 \sqrt{ 73 } ~ \text{and} ~ x = -50+10 \sqrt{ 73 }$.
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