To solve the equation \(2x + 2 = (5 - x) \sqrt{3x - 2}\), we can follow these steps:

1. **Isolate the square root term**:
\[
\frac{2x + 2}{5 - x} = \sqrt{3x - 2}
\]

2. **Square both sides** to eliminate the square root:
\[
\left(\frac{2x + 2}{5 - x}\right)^2 = 3x - 2
\]

3. **Expand and simplify** the left side:
\[
\frac{(2x + 2)^2}{(5 - x)^2} = 3x - 2
\]
\[
\frac{4x^2 + 8x + 4}{25 - 10x + x^2} = 3x - 2
\]

4. **Cross-multiply** to clear the fraction:
\[
4x^2 + 8x + 4 = (3x - 2)(25 - 10x + x^2)
\]

5. **Expand the right side**:
\[
4x^2 + 8x + 4 = 75x - 30x^2 - 20x + 3x^3 - 2x^2
\]
\[
4x^2 + 8x + 4 = 3x^3 - 32x^2 + 55x - 50
\]

6. **Move all terms to one side** to set the equation to zero:
\[
0 = 3x^3 - 32x^2 + 55x - 50 - 4x^2 - 8x - 4
\]
\[
0 = 3x^3 - 36x^2 + 47x - 54
\]

7. **Solve the cubic equation** \(3x^3 - 36x^2 + 47x - 54 = 0\). This can be done using numerical methods or factoring if possible.

Given the complexity of solving cubic equations analytically, it might be more practical to use numerical methods or graphing tools to find the roots. However, we can check for possible rational roots using the Rational Root Theorem, which suggests that any rational root, \( \frac{p}{q} \), is a factor of the constant term (-54) divided by a factor of the leading coefficient (3).

Possible rational roots to test are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm 27, \pm 54 \) divided by \( \pm 1, \pm 3 \).

Testing these values, we find that \( x = 3 \) is a root:

\[
3(3)^3 - 36(3)^2 + 47(3) - 54 = 0
\]

Thus, \( x = 3 \) is a solution. To find other solutions, we can factor \( (x - 3) \) out of the cubic polynomial and solve the resulting quadratic equation.

So, the solution to the equation \(2x + 2 = (5 - x) \sqrt{3x - 2}\) includes \( x = 3 \). Further solutions can be found by solving the remaining quadratic equation.