Find roots of polynomial $$ p(x) = 2999949x^2+200628x+118709 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 33438 }{ 999983 }+0.1961i\\[1 em]x_2 &= -\frac{ 33438 }{ 999983 }-0.1961i \end{aligned} $$Explanation
The solutions of $ 2999949x^2+200628x+118709 = 0 $ are: $ x = -\dfrac{ 33438 }{ 999983 }+0.1961i ~ \text{and} ~ x = -\dfrac{ 33438 }{ 999983 }-0.1961i$.
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