$$2(x+1)(x-4)-4(x-2)(x+2) = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 2(x+1)(x-4)-4(x-2)(x+2) &= 0&& \text{simplify left side} \\[1 em](2x+2)(x-4)-(4x-8)(x+2) &= 0&& \\[1 em]2x^2-8x+2x-8-(4x^2+8x-8x-16) &= 0&& \\[1 em]2x^2-6x-8-(4x^2-16) &= 0&& \\[1 em]2x^2-6x-8-4x^2+16 &= 0&& \\[1 em]-2x^2-6x+8 &= 0&& \\[1 em] \end{aligned} $$
$ -2x^{2}-6x+8 = 0 $ is a quadratic equation.
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