Find roots of polynomial $$ p(x) = b^2-10b+9 $$
Answer
The roots of polynomial $ p(b) $ are:
$$ \begin{aligned}b_1 &= 9\\[1 em]b_2 &= 1 \end{aligned} $$Explanation
The solutions of $ b^2-10b+9 = 0 $ are: $ b = 1 ~ \text{and} ~ b = 9$.
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