Find roots of polynomial $$ p(x) = 3x^2-28x+45 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 7.2701\\[1 em]x_2 &= 2.0633 \end{aligned} $$Explanation
The solutions of $ 3x^2-28x+45 = 0 $ are: $ x = \dfrac{ 14 }{ 3 }-\dfrac{\sqrt{ 61 }}{ 3 } ~ \text{and} ~ x = \dfrac{ 14 }{ 3 }+\dfrac{\sqrt{ 61 }}{ 3 }$.
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