$$4y^2+y-(y^2+2y) = 0$$
Answer
$$ \begin{matrix}y_1 = 0 & y_2 = \dfrac{ 1 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4y^2+y-(y^2+2y) &= 0&& \text{simplify left side} \\[1 em]4y^2+y-y^2-2y &= 0&& \\[1 em]3y^2-y &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{2}-x = 0 } $, first we need to factor our $ x $.
$$ 3x^{2}-x = x \left( 3x-1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ 3x-1 = 0$.
This page was created using
Polynomial Equations Solver