Find roots of polynomial $$ p(x) = 3x^2+13 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{\sqrt{ 39 }}{ 3 }i\\[1 em]x_2 &= - \frac{\sqrt{ 39 }}{ 3 }i \end{aligned} $$Explanation
The solutions of $ 3x^2+13 = 0 $ are: $ x = \dfrac{\sqrt{ 39 }}{ 3 } i ~ \text{and} ~ x = - \dfrac{\sqrt{ 39 }}{ 3 } i $.
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