Find roots of polynomial $$ p(x) = 100x^2+80x+416 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 2 }{ 5 }+2i\\[1 em]x_2 &= -\frac{ 2 }{ 5 }-2i \end{aligned} $$Explanation
The solutions of $ 100x^2+80x+416 = 0 $ are: $ x = -\dfrac{ 2 }{ 5 }+2i ~ \text{and} ~ x = -\dfrac{ 2 }{ 5 }-2i$.
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