$$\frac{1-x^2}{x^2+5x+4} = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = -1 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1-x^2}{x^2+5x+4} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2+5x+4 }. \\[1 em](x^2+5x+4)\frac{1-x^2}{x^2+5x+4} &= (x^2+5x+4)\cdot0&& \text{cancel out the denominators} \\[1 em]1-x^2 &= 0&& \text{simplify left side} \\[1 em]-x^2+1 &= 0&& \\[1 em] \end{aligned} $$
$ -x^{2}+1 = 0 $ is a quadratic equation.
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