Question
$$(x+3)(x\cdot2+4x-2)-(4x\cdot3+3x\cdot2-2x-1) = 0$$
Answer
$$ \begin{matrix}x_1 = - \dfrac{\sqrt{ 30 }}{ 6 } & x_2 = \dfrac{\sqrt{ 30 }}{ 6 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x+3)(x\cdot2+4x-2)-(4x\cdot3+3x\cdot2-2x-1) &= 0&& \text{simplify left side} \\[1 em](x+3)(6x-2)-(12x+6x-2x-1) &= 0&& \\[1 em](x+3)(6x-2)-(16x-1) &= 0&& \\[1 em]6x^2-2x+18x-6-(16x-1) &= 0&& \\[1 em]6x^2+16x-6-(16x-1) &= 0&& \\[1 em]6x^2+16x-6-16x+1 &= 0&& \\[1 em]6x^2+16x-6-16x+1 &= 0&& \\[1 em]6x^2-5 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}-5 = 0 $ is a quadratic equation.
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