Find roots of polynomial $$ p(x) = r^2+9r+18 $$
Answer
The roots of polynomial $ p(r) $ are:
$$ \begin{aligned}r_1 &= -3\\[1 em]r_2 &= -6 \end{aligned} $$Explanation
The solutions of $ r^2+9r+18 = 0 $ are: $ r = -6 ~ \text{and} ~ r = -3$.
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