Question
$$4y(6y-5)+3(6y-5) = 0$$
Answer
$$ \begin{matrix}y_1 = -\dfrac{ 3 }{ 4 } & y_2 = \dfrac{ 5 }{ 6 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4y(6y-5)+3(6y-5) &= 0&& \text{simplify left side} \\[1 em]24y^2-20y+18y-15 &= 0&& \\[1 em]24y^2-2y-15 &= 0&& \\[1 em] \end{aligned} $$
$ 24x^{2}-2x-15 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
This page was created using
Polynomial Equations Solver