To solve the equation \(3x^2 - 3x = (x-1)(x+3)\), we will first expand the right-hand side of the equation.

1. Expand the right-hand side:
\[
(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3
\]

2. Substitute this back into the equation:
\[
3x^2 - 3x = x^2 + 2x - 3
\]

3. Rearrange the equation by bringing all terms to one side:
\[
3x^2 - 3x - x^2 - 2x + 3 = 0
\]

4. Combine like terms:
\[
(3x^2 - x^2) + (-3x - 2x) + 3 = 0
\]
\[
2x^2 - 5x + 3 = 0
\]

5. Now, we will use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -5\), and \(c = 3\):
\[
b^2 - 4ac = (-5)^2 - 4(2)(3) = 25 - 24 = 1
\]

6. Substitute \(b\), \(b^2 - 4ac\), and \(a\) into the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 2} = \frac{5 \pm 1}{4}
\]

7. Solve for the two possible values of \(x\):
- First solution:
\[
x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}
\]
- Second solution:
\[
x = \frac{5 - 1}{4} = \frac{4}{4} = 1
\]

8. The solutions to the equation \(3x^2 - 3x = (x-1)(x+3)\) are:
\[
\boxed{\frac{3}{2}} \quad \text{and} \quad \boxed{1}
\]