Find roots of polynomial $$ p(x) = x^2+20x+148 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -10+4 \sqrt{ 3 }i\\[1 em]x_2 &= -10-4 \sqrt{ 3 }i \end{aligned} $$Explanation
The solutions of $ x^2+20x+148 = 0 $ are: $ x = -10+4 \sqrt{ 3 }i ~ \text{and} ~ x = -10-4 \sqrt{ 3 }i$.
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