$$(3x-1)\cdot(1-6x) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = \dfrac{ 1 }{ 6 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-1)\cdot(1-6x) &= 0&& \text{simplify left side} \\[1 em]3x-18x^2-1+6x &= 0&& \\[1 em]-18x^2+9x-1 &= 0&& \\[1 em] \end{aligned} $$
$ -18x^{2}+9x-1 = 0 $ is a quadratic equation.
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