Question
Find roots of polynomial $$ p(x) = 3c^2-12c+14 $$
Answer
The roots of polynomial $ p(c) $ are:
$$ \begin{aligned}c_1 &= 2+\dfrac{\sqrt{ 6 }}{ 3 }i\\[1 em]c_2 &= 2- \dfrac{\sqrt{ 6 }}{ 3 }i \end{aligned} $$Explanation
The solutions of $ 3c^2-12c+14 = 0 $ are: $ c = 2+\dfrac{\sqrt{ 6 }}{ 3 }i ~ \text{and} ~ c = 2-\dfrac{\sqrt{ 6 }}{ 3 }i$.
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