Find roots of polynomial $$ p(x) = 10x^2+x-8 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 0.8458\\[1 em]x_2 &= -0.9458 \end{aligned} $$Explanation
The solutions of $ 10x^2+x-8 = 0 $ are: $ x = -\dfrac{ 1 }{ 20 }-\dfrac{\sqrt{ 321 }}{ 20 } ~ \text{and} ~ x = -\dfrac{ 1 }{ 20 }+\dfrac{\sqrt{ 321 }}{ 20 }$.
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