To solve the equation

\[
|2x + 3| + |2x - 1| = \frac{8}{2(y - 5)^2 + 2}
\]

we first simplify the right-hand side:

The expression \(2(y - 5)^2 + 2\) can be rewritten as:

\[
2((y - 5)^2 + 1)
\]

So,

\[
\frac{8}{2(y - 5)^2 + 2} = \frac{8}{2((y - 5)^2 + 1)} = \frac{4}{(y - 5)^2 + 1}
\]

Now the equation becomes:

\[
|2x + 3| + |2x - 1| = \frac{4}{(y - 5)^2 + 1}
\]

### Step 1: Solve the left side

Next, we need to analyze the left side, \( |2x + 3| + |2x - 1| \). We’ll consider different cases based on the values of \(2x + 3\) and \(2x - 1\).

**Case 1:** \(x < -\frac{3}{2}\)
In this case, \(2x + 3 < 0\) and \(2x - 1 < 0\).

\[
|2x + 3| + |2x - 1| = -(2x + 3) - (2x - 1) = -2x - 3 - 2x + 1 = -4x - 2
\]

**Case 2:** \(-\frac{3}{2} \leq x < \frac{1}{2}\)
Here, \(2x + 3 \geq 0\) and \(2x - 1 < 0\).

\[
|2x + 3| + |2x - 1| = (2x + 3) - (2x - 1) = 2x + 3 - 2x + 1 = 4
\]

**Case 3:** \(x \geq \frac{1}{2}\)
In this case, \(2x + 3 \geq 0\) and \(2x - 1 \geq 0\).

\[
|2x + 3| + |2x - 1| = (2x + 3) + (2x - 1) = 4x + 2
\]

### Step 2: Setting up the equation

Now we set up equations for each case:

1. **Case 1:**
\(-4x - 2 = \frac{4}{(y - 5)^2 + 1}\)

2. **Case 2:**
\(4 = \frac{4}{(y - 5)^2 + 1}\)

3. **Case 3:**
\(4x + 2 = \frac{4}{(y - 5)^2 + 1}\)

### Step 3: Solve each case

**Case 2** is the simplest:

\[
4 = \frac{4}{(y - 5)^2 + 1}
\]

Cross-multiplying gives:

\[
4((y - 5)^2 + 1) = 4 \implies (y - 5)^2 + 1 = 1 \implies (y - 5)^2 = 0 \implies y = 5
\]

For **Case 1** and **Case 3**, we can set similar equations and solve for \(x\) in those respective ranges.

### Conclusion

The solution for \(y\) is \(5\). Depending on the range of \(x\), you can find corresponding values for \(x\) by solving the linear equations derived for each case.