Find roots of polynomial $$ p(x) = x^2-8x+97 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= 4+9i\\[1 em]x_2 &= 4-9i \end{aligned} $$Explanation
The solutions of $ x^2-8x+97 = 0 $ are: $ x = 4+9i ~ \text{and} ~ x = 4-9i$.
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