$$(3-6x)^2-(2x+8)^2 = 0$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 5 }{ 8 } & x_2 = \dfrac{ 11 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3-6x)^2-(2x+8)^2 &= 0&& \text{simplify left side} \\[1 em]9-36x+36x^2-(4x^2+32x+64) &= 0&& \\[1 em]9-36x+36x^2-4x^2-32x-64 &= 0&& \\[1 em]32x^2-68x-55 &= 0&& \\[1 em] \end{aligned} $$
$ 32x^{2}-68x-55 = 0 $ is a quadratic equation.
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