Find roots of polynomial $$ p(x) = 10x^2-40x-11 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4.2583\\[1 em]x_2 &= -0.2583 \end{aligned} $$Explanation
The solutions of $ 10x^2-40x-11 = 0 $ are: $ x = 2-\dfrac{\sqrt{ 510 }}{ 10 } ~ \text{and} ~ x = 2+\dfrac{\sqrt{ 510 }}{ 10 }$.
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