Three, (16 points) Let the equation $x^{2}-|2 x-1|-4=0$. Find the sum of all roots that satisfy the equation.


When $2 x-1>0$, i.e., $x>\frac{1}{2}$, the original equation becomes $x^{2}-2 x-3=0$.
Solving, we get $x_{1}=3, x_{2}=-1$ (discard).
When $2 x-1=0$, i.e., $x=\frac{1}{2}$, substituting into the original equation does not fit, discard.
When $2 x-1<0$, i.e., $x<\frac{1}{2}$, the original equation becomes $x^{2}+2 x-5=0$.
Solving, we get $x_{1}=-1+\sqrt{6}$ (discard), $x_{2}=-1-\sqrt{6}$.
So, all roots of the equation are 3 and $-1-\sqrt{6}$.
Therefore, the sum of all roots is $3+(-1-\sqrt{6})=2-\sqrt{6}$.