Find roots of polynomial $$ p(x) = x^2-21x+100 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 13.7016\\[1 em]x_2 &= 7.2984 \end{aligned} $$Explanation
The solutions of $ x^2-21x+100 = 0 $ are: $ x = \dfrac{ 21 }{ 2 }-\dfrac{\sqrt{ 41 }}{ 2 } ~ \text{and} ~ x = \dfrac{ 21 }{ 2 }+\dfrac{\sqrt{ 41 }}{ 2 }$.
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