$$(9c+5)(9c-5) = 0$$
Answer
$$ \begin{matrix}c_1 = \dfrac{ 5 }{ 9 } & c_2 = -\dfrac{ 5 }{ 9 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (9c+5)(9c-5) &= 0&& \text{simplify left side} \\[1 em]81c^2-45c+45c-25 &= 0&& \\[1 em]81c^2-45c+45c-25 &= 0&& \\[1 em]81c^2-25 &= 0&& \\[1 em] \end{aligned} $$
$ 81x^{2}-25 = 0 $ is a quadratic equation.
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