Find roots of polynomial $$ p(x) = 10y^2+80y+200 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= -4+2i\\[1 em]y_2 &= -4-2i \end{aligned} $$Explanation
The solutions of $ 10y^2+80y+200 = 0 $ are: $ y = -4+2i ~ \text{and} ~ y = -4-2i$.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this quadratic equation.
This page was created using
Polynomial Roots Calculator