Question
$$(3x-1)(2x\cdot3+4x\cdot2-5) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = \dfrac{ 5 }{ 14 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-1)(2x\cdot3+4x\cdot2-5) &= 0&& \text{simplify left side} \\[1 em](3x-1)(6x+8x-5) &= 0&& \\[1 em](3x-1)(14x-5) &= 0&& \\[1 em]42x^2-15x-14x+5 &= 0&& \\[1 em]42x^2-29x+5 &= 0&& \\[1 em] \end{aligned} $$
$ 42x^{2}-29x+5 = 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
Equations Solver