The quadratic polynomial $f(x)$ has the expansion $2x^2 - 3x + r$. What is the largest real value of $r$ for which the ranges of the functions $f(x)$ and $f(f(x))$ are the same set?
1. **Determine the range of \( f(x) \):**
   The given quadratic polynomial is \( f(x) = 2x^2 - 3x + r \). To find the range of \( f(x) \), we first find its vertex. The vertex form of a quadratic function \( ax^2 + bx + c \) is given by the vertex \((h, k)\), where \( h = -\frac{b}{2a} \).

   For \( f(x) = 2x^2 - 3x + r \):
   \[
   h = -\frac{-3}{2 \cdot 2} = \frac{3}{4}
   \]

   Now, substitute \( x = \frac{3}{4} \) into \( f(x) \) to find the minimum value (since the parabola opens upwards, the vertex gives the minimum value):
   \[
   f\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + r = 2 \cdot \frac{9}{16} - \frac{9}{4} + r = \frac{18}{16} - \frac{36}{16} + r = -\frac{18}{16} + r = -\frac{9}{8} + r
   \]

   Therefore, the range of \( f(x) \) is:
   \[
   f(\mathbb{R}) = \left[ r - \frac{9}{8}, \infty \right)
   \]

2. **Determine the range of \( f(f(x)) \):**
   To ensure that \( f(f(x)) \subseteq f(\mathbb{R}) \), we need to check the minimum value of \( f(f(x)) \). For \( f(f(x)) \) to have the same range as \( f(x) \), the minimum value of \( f(f(x)) \) must be at least \( r - \frac{9}{8} \).

   Consider \( f(f(x)) \):
   \[
   f(f(x)) = f(2x^2 - 3x + r) = 2(2x^2 - 3x + r)^2 - 3(2x^2 - 3x + r) + r
   \]

   To simplify the problem, we need to ensure that the minimum value of \( f(f(x)) \) is within the range of \( f(x) \). The minimum value of \( f(x) \) is \( r - \frac{9}{8} \). Therefore, we need:
   \[
   f\left(\frac{3}{4}\right) = r - \frac{9}{8} \leq 3/4
   \]

3. **Solve the inequality:**
   \[
   r - \frac{9}{8} \leq \frac{3}{4}
   \]
   \[
   r \leq \frac{3}{4} + \frac{9}{8}
   \]
   \[
   r \leq \frac{6}{8} + \frac{9}{8}
   \]
   \[
   r \leq \frac{15}{8}
   \]

   Hence, the largest real value of \( r \) for which the ranges of the functions \( f(x) \) and \( f(f(x)) \) are the same set is:
   \[
   r_{\text{max}} = \frac{15}{8}
   \]

The final answer is \( \boxed{ \frac{15}{8} } \)