$$\frac{4x+8}{x^2-4} = 0$$

$$ \begin{aligned} \frac{4x+8}{x^2-4} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2-4 }. \\[1 em](x^2-4)\frac{4x+8}{x^2-4} &= (x^2-4)\cdot0&& \text{cancel out the denominators} \\[1 em]4x+8 &= 0&& \text{ move the constants to the right } \\[1 em]4x &= -8&& \text{ divide both sides by $ 4 $ } \\[1 em]x &= -\frac{8}{4}&& \\[1 em]x &= -2&& \\[1 em] \end{aligned} $$

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