$$(x-4)(2x+5x-3) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 3 }{ 7 } & x_2 = 4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x-4)(2x+5x-3) &= 0&& \text{simplify left side} \\[1 em](x-4)(7x-3) &= 0&& \\[1 em]7x^2-3x-28x+12 &= 0&& \\[1 em]7x^2-31x+12 &= 0&& \\[1 em] \end{aligned} $$
$ 7x^{2}-31x+12 = 0 $ is a quadratic equation.
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