Question
$$(4-(3c-1))\cdot(6-(3c-1)) = 0$$
Answer
$$ \begin{matrix}c_1 = \dfrac{ 5 }{ 3 } & c_2 = \dfrac{ 7 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (4-(3c-1))\cdot(6-(3c-1)) &= 0&& \text{simplify left side} \\[1 em](4-3c+1)(6-3c+1) &= 0&& \\[1 em](-3c+5)(-3c+7) &= 0&& \\[1 em]9c^2-21c-15c+35 &= 0&& \\[1 em]9c^2-36c+35 &= 0&& \\[1 em] \end{aligned} $$
$ 9x^{2}-36x+35 = 0 $ is a quadratic equation.
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