Find roots of polynomial $$ p(x) = 2y^2-13y-45 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 9\\[1 em]y_2 &= -\frac{ 5 }{ 2 } \end{aligned} $$Explanation
The solutions of $ 2y^2-13y-45 = 0 $ are: $ y = -\dfrac{ 5 }{ 2 } ~ \text{and} ~ y = 9$.
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