Find roots of polynomial $$ p(x) = 2x^2-7x+1 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 3.3508\\[1 em]x_2 &= 0.1492 \end{aligned} $$Explanation
The solutions of $ 2x^2-7x+1 = 0 $ are: $ x = \dfrac{ 7 }{ 4 }-\dfrac{\sqrt{ 41 }}{ 4 } ~ \text{and} ~ x = \dfrac{ 7 }{ 4 }+\dfrac{\sqrt{ 41 }}{ 4 }$.
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