Question
$$3(4x-1)\cdot3(x\cdot2+1)\cdot4-(4x-1)\cdot4(x\cdot2+1)\cdot3 = 0$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 1 }{ 2 } & x_2 = \dfrac{ 1 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3(4x-1)\cdot3(x\cdot2+1)\cdot4-(4x-1)\cdot4(x\cdot2+1)\cdot3 &= 0&& \text{simplify left side} \\[1 em](12x-3)\cdot3(x\cdot2+1)\cdot4-(16x-4)(x\cdot2+1)\cdot3 &= 0&& \\[1 em](36x-9)(x\cdot2+1)\cdot4-(32x^2+16x-8x-4)\cdot3 &= 0&& \\[1 em](36x-9)(x\cdot2+1)\cdot4-(32x^2+8x-4)\cdot3 &= 0&& \\[1 em](72x^2+36x-18x-9)\cdot4-(96x^2+24x-12) &= 0&& \\[1 em](72x^2+18x-9)\cdot4-(96x^2+24x-12) &= 0&& \\[1 em]288x^2+72x-36-(96x^2+24x-12) &= 0&& \\[1 em]288x^2+72x-36-96x^2-24x+12 &= 0&& \\[1 em]192x^2+48x-24 &= 0&& \\[1 em] \end{aligned} $$
$ 192x^{2}+48x-24 = 0 $ is a quadratic equation.
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