Find roots of polynomial $$ p(x) = 3x^2+19x+2004 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -\frac{ 19 }{ 6 }+\frac{\sqrt{ 23687 }}{ 6 }i\\[1 em]x_2 &= -\frac{ 19 }{ 6 }- \frac{\sqrt{ 23687 }}{ 6 }i \end{aligned} $$Explanation
The solutions of $ 3x^2+19x+2004 = 0 $ are: $ x = -\dfrac{ 19 }{ 6 }+\dfrac{\sqrt{ 23687 }}{ 6 }i ~ \text{and} ~ x = -\dfrac{ 19 }{ 6 }-\dfrac{\sqrt{ 23687 }}{ 6 }i$.
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