Question
Find roots of polynomial $$ p(x) = x^2-96x+288 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 92.8999\\[1 em]x_2 &= 3.1001 \end{aligned} $$Explanation
The solutions of $ x^2-96x+288 = 0 $ are: $ x = 48-12 \sqrt{ 14 } ~ \text{and} ~ x = 48+12 \sqrt{ 14 }$.
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