Question
Find roots of polynomial $$ p(x) = 16c^2-9 $$
Answer
The roots of polynomial $ p(c) $ are:
$$ \begin{aligned}c_1 &= \dfrac{ 3 }{ 4 }\\[1 em]c_2 &= -\dfrac{ 3 }{ 4 } \end{aligned} $$Explanation
The solutions of $ 16c^2-9 = 0 $ are: $ c = -\dfrac{ 3 }{ 4 } ~ \text{and} ~ c = \dfrac{ 3 }{ 4 }$.
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