Question
Find roots of polynomial $$ p(x) = 61x^2-38x-19 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.9506\\[1 em]x_2 &= -0.3277 \end{aligned} $$Explanation
The solutions of $ 61x^2-38x-19 = 0 $ are: $ x = \dfrac{ 19 }{ 61 }-\dfrac{ 4 \sqrt{ 95}}{ 61 } ~ \text{and} ~ x = \dfrac{ 19 }{ 61 }+\dfrac{ 4 \sqrt{ 95}}{ 61 }$.
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