Find roots of polynomial $$ p(x) = r^2+6r+25 $$
Answer
The roots of polynomial $ p(r) $ are:
$$ \begin{aligned}r_1 &= -3+4i\\[1 em]r_2 &= -3-4i \end{aligned} $$Explanation
The solutions of $ r^2+6r+25 = 0 $ are: $ r = -3+4i ~ \text{and} ~ r = -3-4i$.
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