Find roots of polynomial $$ p(x) = y^2-3y-4 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 4\\[1 em]y_2 &= -1 \end{aligned} $$Explanation
The solutions of $ y^2-3y-4 = 0 $ are: $ y = -1 ~ \text{and} ~ y = 4$.
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