Find roots of polynomial $$ p(x) = x^2-408x+1084 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 405.3256\\[1 em]x_2 &= 2.6744 \end{aligned} $$Explanation
The solutions of $ x^2-408x+1084 = 0 $ are: $ x = 204-2 \sqrt{ 10133 } ~ \text{and} ~ x = 204+2 \sqrt{ 10133 }$.
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