Find roots of polynomial $$ p(x) = 3x^2-200x+4900 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 100 }{ 3 }+\frac{ 10 \sqrt{ 47}}{ 3 }i\\[1 em]x_2 &= \frac{ 100 }{ 3 }-10 \frac{\sqrt{ 47 }}{ 3 }i \end{aligned} $$Explanation
The solutions of $ 3x^2-200x+4900 = 0 $ are: $ x = \dfrac{ 100 }{ 3 }+\dfrac{ 10 \sqrt{ 47}}{ 3 }i ~ \text{and} ~ x = \dfrac{ 100 }{ 3 }-\dfrac{ 10 \sqrt{ 47}}{ 3 }i$.
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