$$(x-\frac{1}{4})(x+3) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 4 } & x_2 = -3 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x-\frac{1}{4})(x+3) &= 0&& \text{simplify left side} \\[1 em]\frac{4x-1}{4}(x+3) &= 0&& \\[1 em]\frac{4x^2+11x-3}{4} &= 0&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{4x^2+11x-3}{4} &= 4\cdot0&& \text{cancel out the denominators} \\[1 em]4x^2+11x-3 &= 0&& \\[1 em] \end{aligned} $$
$ 4x^{2}+11x-3 = 0 $ is a quadratic equation.
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