Question
Find roots of polynomial $$ p(x) = 29x^2-297x+900 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \dfrac{ 297 }{ 58 }+\dfrac{ 3 \sqrt{ 1799}}{ 58 }i\\[1 em]x_2 &= \dfrac{ 297 }{ 58 }-3 \dfrac{\sqrt{ 1799 }}{ 58 }i \end{aligned} $$Explanation
The solutions of $ 29x^2-297x+900 = 0 $ are: $ x = \dfrac{ 297 }{ 58 }+\dfrac{ 3 \sqrt{ 1799}}{ 58 }i ~ \text{and} ~ x = \dfrac{ 297 }{ 58 }-\dfrac{ 3 \sqrt{ 1799}}{ 58 }i$.
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