Let \(P(x)=x^2+\dfrac x 2 +b\) and \(Q(x)=x^2+cx+d\) be two polynomials with real coefficients such that \(P(x)Q(x)=Q(P(x))\) for all real \(x\). Find all real roots of \(P(Q(x))=0\).
1. **Given Polynomials:**
   \[
   P(x) = x^2 + \frac{x}{2} + b \quad \text{and} \quad Q(x) = x^2 + cx + d
   \]
   We are given that \(P(x)Q(x) = Q(P(x))\) for all real \(x\).

2. **Setting \(P(x) = 0\):**
   \[
   P(x) = 0 \implies x^2 + \frac{x}{2} + b = 0
   \]
   Let the roots of \(P(x) = 0\) be \(\alpha\) and \(\beta\).

3. **Evaluating \(Q(0)\):**
   \[
   P(0) = b \implies Q(0) = d
   \]
   Since \(P(x)Q(x) = Q(P(x))\), substituting \(x = 0\) gives:
   \[
   P(0)Q(0) = Q(P(0)) \implies b \cdot d = Q(b)
   \]
   If \(P(x) = 0\) has real roots, then \(Q(0) = 0\) implies \(d = 0\).

4. **Expanding Both Sides:**
   \[
   P(x)Q(x) = (x^2 + \frac{x}{2} + b)(x^2 + cx + d)
   \]
   \[
   Q(P(x)) = Q(x^2 + \frac{x}{2} + b) = (x^2 + \frac{x}{2} + b)^2 + c(x^2 + \frac{x}{2} + b) + d
   \]
   Since \(d = 0\), we simplify:
   \[
   P(x)Q(x) = (x^2 + \frac{x}{2} + b)(x^2 + cx)
   \]
   \[
   Q(P(x)) = (x^2 + \frac{x}{2} + b)^2 + c(x^2 + \frac{x}{2} + b)
   \]

5. **Equating Coefficients:**
   By comparing coefficients of \(x^4, x^3, x^2, x, \text{and constant terms}\) on both sides, we get:
   \[
   x^4: 1 = 1
   \]
   \[
   x^3: c + \frac{1}{2} = c + \frac{1}{2}
   \]
   \[
   x^2: b + \frac{c}{2} = b + \frac{c}{2}
   \]
   \[
   x: \frac{b}{2} = \frac{b}{2}
   \]
   \[
   \text{constant term: } b^2 + cb = b^2 + cb
   \]
   These equations are consistent, confirming our values.

6. **Finding Roots of \(P(Q(x)) = 0\):**
   \[
   P(Q(x)) = 0 \implies Q(x) = \alpha \text{ or } \beta
   \]
   Since \(Q(x) = x^2 + cx\) and \(d = 0\), we solve:
   \[
   Q(x) = -1 \implies x^2 + cx + 1 = 0
   \]
   The discriminant \(\Delta = c^2 - 4\) must be non-negative for real roots:
   \[
   c^2 - 4 < 0 \implies \text{no real roots}
   \]
   \[
   Q(x) = \frac{1}{2} \implies x^2 + cx - \frac{1}{2} = 0
   \]
   The discriminant \(\Delta = c^2 + 2\) is always positive, giving real roots:
   \[
   x = \frac{-c \pm \sqrt{c^2 + 2}}{2}
   \]

The final answer is \( \boxed{ x = \frac{-c \pm \sqrt{c^2 + 2}}{2} } \)