Find roots of polynomial $$ p(x) = x^2-9x+36 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 9 }{ 2 }+\frac{ 3 \sqrt{ 7}}{ 2 }i\\[1 em]x_2 &= \frac{ 9 }{ 2 }-3 \frac{\sqrt{ 7 }}{ 2 }i \end{aligned} $$Explanation
The solutions of $ x^2-9x+36 = 0 $ are: $ x = \dfrac{ 9 }{ 2 }+\dfrac{ 3 \sqrt{ 7}}{ 2 }i ~ \text{and} ~ x = \dfrac{ 9 }{ 2 }-\dfrac{ 3 \sqrt{ 7}}{ 2 }i$.
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