$$(2-z)(4z-1) = 0$$
Answer
$$ \begin{matrix}z_1 = \dfrac{ 1 }{ 4 } & z_2 = 2 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (2-z)(4z-1) &= 0&& \text{simplify left side} \\[1 em]8z-2-4z^2+z &= 0&& \\[1 em]-4z^2+9z-2 &= 0&& \\[1 em] \end{aligned} $$
$ -4x^{2}+9x-2 = 0 $ is a quadratic equation.
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