Find roots of polynomial $$ p(x) = x^2+10x+41 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -5+4i\\[1 em]x_2 &= -5-4i \end{aligned} $$Explanation
The solutions of $ x^2+10x+41 = 0 $ are: $ x = -5+4i ~ \text{and} ~ x = -5-4i$.
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