Question
$$(4x-2)^2-4(4x-2)+3 = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 3 }{ 4 } & x_2 = \dfrac{ 5 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (4x-2)^2-4(4x-2)+3 &= 0&& \text{simplify left side} \\[1 em]16x^2-16x+4-4(4x-2)+3 &= 0&& \\[1 em]16x^2-16x+4-(16x-8)+3 &= 0&& \\[1 em]16x^2-16x+4-16x+8+3 &= 0&& \\[1 em]16x^2-32x+15 &= 0&& \\[1 em] \end{aligned} $$
$ 16x^{2}-32x+15 = 0 $ is a quadratic equation.
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