Find roots of polynomial $$ p(x) = 10b^2-27b+18 $$
Answer
The roots of polynomial $ p(b) $ are:
$$ \begin{aligned}b_1 &= \frac{ 3 }{ 2 }\\[1 em]b_2 &= \frac{ 6 }{ 5 } \end{aligned} $$Explanation
The solutions of $ 10b^2-27b+18 = 0 $ are: $ b = \dfrac{ 6 }{ 5 } ~ \text{and} ~ b = \dfrac{ 3 }{ 2 }$.
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