Find roots of polynomial $$ p(x) = 41x^2+142x-49 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.3162\\[1 em]x_2 &= -3.7796 \end{aligned} $$Explanation
The solutions of $ 41x^2+142x-49 = 0 $ are: $ x = -\dfrac{ 71 }{ 41 }-\dfrac{ 5 \sqrt{ 282}}{ 41 } ~ \text{and} ~ x = -\dfrac{ 71 }{ 41 }+\dfrac{ 5 \sqrt{ 282}}{ 41 }$.
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