$$\frac{4}{x^2+2x-4} = \frac{4}{x+4}$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 1 }{ 2 }-\dfrac{\sqrt{ 33 }}{ 2 } & x_2 = -\dfrac{ 1 }{ 2 }+\dfrac{\sqrt{ 33 }}{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{4}{x^2+2x-4} &= \frac{4}{x+4}&& \text{multiply ALL terms by } \color{blue}{ (x^2+2x-4)(x+4) }. \\[1 em](x^2+2x-4)(x+4)\cdot\frac{4}{x^2+2x-4} &= (x^2+2x-4)(x+4)\cdot\frac{4}{x+4}&& \text{cancel out the denominators} \\[1 em]4x+16 &= 4x^2+8x-16&& \text{move all terms to the left hand side } \\[1 em]4x+16-4x^2-8x+16 &= 0&& \text{simplify left side} \\[1 em]-4x^2-4x+32 &= 0&& \\[1 em] \end{aligned} $$
$ -4x^{2}-4x+32 = 0 $ is a quadratic equation.
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