Question
$$\frac{2}{x^2-x-20}+\frac{3}{x^2+7x+12} = 0$$
Answer
$$ \begin{matrix}x_1 = -4 & x_2 = \dfrac{ 9 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{2}{x^2-x-20}+\frac{3}{x^2+7x+12} &= 0&& \text{multiply ALL terms by } \color{blue}{ (x^2-x-20)(x^2+7x+12) }. \\[1 em](x^2-x-20)(x^2+7x+12)\cdot\frac{2}{x^2-x-20}+(x^2-x-20)(x^2+7x+12)\cdot\frac{3}{x^2+7x+12} &= (x^2-x-20)(x^2+7x+12)\cdot0&& \text{cancel out the denominators} \\[1 em]2x^2+14x+24+3x^2-3x-60 &= 0&& \text{simplify left side} \\[1 em]5x^2+11x-36 &= 0&& \\[1 em] \end{aligned} $$
$ 5x^{2}+11x-36 = 0 $ is a quadratic equation.
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