Find roots of polynomial $$ p(x) = x^2+28x-1040 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 21.1568\\[1 em]x_2 &= -49.1568 \end{aligned} $$Explanation
The solutions of $ x^2+28x-1040 = 0 $ are: $ x = -14-2 \sqrt{ 309 } ~ \text{and} ~ x = -14+2 \sqrt{ 309 }$.
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