I want to understand the equation
$$|x+1|=x^2 -1$$ $$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$ $$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \Rightarrow x^2 + x = 0 \Rightarrow x_3= 0 \text{, (but doesn't fulfill } x+1 < 0), x_4 = -1$$
$$\Rightarrow L = \{2,-1\}$$
What I don't get is how $x+1 < 0 \Rightarrow x^2 + x = 0$. How do we get $x^2 + x = 0$?
Suppose that $x$ satisfies
$$(*) \quad |x+1|=x^2 -1.$$
Then $|x+1|=(x -1)(x+1).$ Hence $|x+1|=|x-1| \cdot |x+1|.$
It is clear that $x=-1$ is a solution of $(*)$. Now we assume that $x \ne -1.$ Then we get $|x-1|=1.$ The last equation has the solutions $x=0$ and $x=2$. But only $x=2$ is a solutions of $(*)$.
Consequence: $(*)$ holds $ \iff x \in \{-1,2\}.$
When $x + 1 < 0$ then the initial equation becomes $$\begin{align*} && -(x+1) &= x^2 - 1 \\ &\iff&-x -1 &= x^2 - 1 \\ &\iff& 0 &= x^2 + x\end{align*}$$
For $x+1 \geq 0$, $|x+1| = x+1$, and hence the equation reads as:
$$x+1-x^2 +1 = 0 \Rightarrow x^2 -x - 2=0.$$
For $x+1 < 0$, $|x+1| = -(x+1)$, and hence the equation reads as:
$$-(x+1)-x^2 +1 = 0 \Rightarrow x^2 +x=0.$$
