Find roots of polynomial $$ p(x) = x^2+19x+90 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= -9\\[1 em]x_2 &= -10 \end{aligned} $$Explanation
The solutions of $ x^2+19x+90 = 0 $ are: $ x = -10 ~ \text{and} ~ x = -9$.
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