$$\frac{x}{x+2} = -\frac{2}{(x+1)(x+2)}$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 3 }{ 2 }+\dfrac{\sqrt{ 15 }}{ 6 }i & x_2 = -\dfrac{ 3 }{ 2 }-\dfrac{\sqrt{ 15 }}{ 6 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x}{x+2} &= -\frac{2}{(x+1)(x+2)}&& \text{multiply ALL terms by } \color{blue}{ (x+2)(x+1) }. \\[1 em](x+2)(x+1)\frac{x}{x+2} &= -(x+2)(x+1)\cdot\frac{2}{(x+1)(x+2)}&& \text{cancel out the denominators} \\[1 em]x^2+x &= -(2x^2+8x+8)&& \text{simplify right side} \\[1 em]x^2+x &= -2x^2-8x-8&& \text{move all terms to the left hand side } \\[1 em]x^2+x+2x^2+8x+8 &= 0&& \text{simplify left side} \\[1 em]3x^2+9x+8 &= 0&& \\[1 em] \end{aligned} $$
$ 3x^{2}+9x+8 = 0 $ is a quadratic equation.
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