Find the number of real numbers which satisfy the equation $x|x-1|-4|x|+3=0$.

$ \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4 $
To solve the equation \( x|x-1| - 4|x| + 3 = 0 \), we need to consider different cases based on the value of \( x \) because of the absolute value functions. We will split the problem into three cases: \( x \geq 1 \), \( 0 \leq x < 1 \), and \( x < 0 \).

1. **Case 1: \( x \geq 1 \)**
   - Here, \( |x-1| = x-1 \) and \( |x| = x \).
   - Substituting these into the equation, we get:
     \[
     x(x-1) - 4x + 3 = 0
     \]
     Simplifying, we have:
     \[
     x^2 - x - 4x + 3 = 0 \implies x^2 - 5x + 3 = 0
     \]
   - To find the roots, we use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     x = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}
     \]
   - The roots are:
     \[
     x = \frac{5 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{5 - \sqrt{13}}{2}
     \]
   - Since \( x \geq 1 \), we need to check which of these roots satisfy this condition:
     \[
     \frac{5 + \sqrt{13}}{2} \approx 4.302 \quad \text{(valid)}
     \]
     \[
     \frac{5 - \sqrt{13}}{2} \approx 0.698 \quad \text{(not valid)}
     \]
   - Therefore, there is 1 valid solution in this case.

2. **Case 2: \( 0 \leq x < 1 \)**
   - Here, \( |x-1| = 1-x \) and \( |x| = x \).
   - Substituting these into the equation, we get:
     \[
     x(1-x) - 4x + 3 = 0
     \]
     Simplifying, we have:
     \[
     x - x^2 - 4x + 3 = 0 \implies -x^2 - 3x + 3 = 0 \implies x^2 + 3x - 3 = 0
     \]
   - To find the roots, we use the quadratic formula:
     \[
     x = \frac{-3 \pm \sqrt{9 + 12}}{2} = \frac{-3 \pm \sqrt{21}}{2}
     \]
   - The roots are:
     \[
     x = \frac{-3 + \sqrt{21}}{2} \quad \text{and} \quad x = \frac{-3 - \sqrt{21}}{2}
     \]
   - Since \( 0 \leq x < 1 \), we need to check which of these roots satisfy this condition:
     \[
     \frac{-3 + \sqrt{21}}{2} \approx 0.791 \quad \text{(valid)}
     \]
     \[
     \frac{-3 - \sqrt{21}}{2} \approx -3.791 \quad \text{(not valid)}
     \]
   - Therefore, there is 1 valid solution in this case.

3. **Case 3: \( x < 0 \)**
   - Here, \( |x-1| = 1-x \) and \( |x| = -x \).
   - Substituting these into the equation, we get:
     \[
     x(1-x) + 4x + 3 = 0
     \]
     Simplifying, we have:
     \[
     x - x^2 + 4x + 3 = 0 \implies x^2 + 3x + 3 = 0
     \]
   - To find the roots, we use the quadratic formula:
     \[
     x = \frac{-3 \pm \sqrt{9 - 12}}{2} = \frac{-3 \pm \sqrt{-3}}{2}
     \]
   - Since the discriminant is negative (\(\Delta = -3\)), there are no real solutions in this case.

In total, we have \(1\) solution from Case 1 and \(1\) solution from Case 2, giving us a total of \(2\) real solutions.

The final answer is \(\boxed{2}\)