Find roots of polynomial $$ p(x) = x^2-20x+90 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 13.1623\\[1 em]x_2 &= 6.8377 \end{aligned} $$Explanation
The solutions of $ x^2-20x+90 = 0 $ are: $ x = 10-\sqrt{ 10 } ~ \text{and} ~ x = 10+\sqrt{ 10 }$.
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