Question
Find roots of polynomial $$ p(x) = a^2-4a+16 $$
Answer
The roots of polynomial $ p(a) $ are:
$$ \begin{aligned}a_1 &= 2+2 \sqrt{ 3 }i\\[1 em]a_2 &= 2-2 \sqrt{ 3 }i \end{aligned} $$Explanation
The solutions of $ a^2-4a+16 = 0 $ are: $ a = 2+2 \sqrt{ 3 }i ~ \text{and} ~ a = 2-2 \sqrt{ 3 }i$.
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