Find roots of polynomial $$ p(x) = 3x^2-23x+12 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 7.1036\\[1 em]x_2 &= 0.5631 \end{aligned} $$Explanation
The solutions of $ 3x^2-23x+12 = 0 $ are: $ x = \dfrac{ 23 }{ 6 }-\dfrac{\sqrt{ 385 }}{ 6 } ~ \text{and} ~ x = \dfrac{ 23 }{ 6 }+\dfrac{\sqrt{ 385 }}{ 6 }$.
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