The equation $|x^2-2x-3|=b$ then for what values of $\text{'}b\text{'}$ there are three solutions?

$x^2-2x-3=0\Rightarrow x=3,-1$

$(-\frac{b}{2a},-\frac{\Delta }{4a})=(1,-4)$

$y=|x^2-2x-3|=b$

Only 2 solutions: $y=b=0$ and $y=b>4$

4 solutions: $0<b<4$

No solution: $y=b<0$