Find roots of polynomial $$ p(x) = -y^2-10y+100 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 6.1803\\[1 em]y_2 &= -16.1803 \end{aligned} $$Explanation
The solutions of $ -y^2-10y+100 = 0 $ are: $ y = -5-5 \sqrt{ 5 } ~ \text{and} ~ y = -5+5 \sqrt{ 5 }$.
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