I think answer : $ D $

Solving for them we have that $x=\frac{5-\sqrt{37}}{2}, \frac{5+\sqrt{37}}{2}, \frac{\sqrt{21}-3}{2}$, hence the answer is $\textbf{(D) }3$

Full Solution: As in most typical absolute value problems, we split the value of $x$ into three intervals: (i) $x\ge1$, (ii) $1>x\ge 0$, and (iii) $x<0$. Case (i): If $x\ge 1$, then our equation becomes $x(x-1)-4x+3=x^2-5x+3=0$, and since the discriminant is $\Delta=25-4\cdot3=13$, there are $2$ real solutions in this case. $\Box$ Case (ii): If $1>x\ge0$, our equation is $x(1-x)-4x+3=x-x^2-4x+3=-x^2-3x+3$. We have $2$ real roots, but one is less than zero, so we only have $1$ solution for this case. $\Box$ Case (iii): If $x<0$, we have $x(x-1)+4x+3=x^2+3x+3=0$. Our discriminant is $\Delta=9-12=-3$, so there are no real solutions in this case. $\Box$ In total we have $3$ solutions.