Find all the real roots of the system of equations:
$$ \begin{cases} x^3+y^3=19 \\ x^2+y^2+5x+5y+xy=12 \end{cases}  $$
To solve the system of equations:
\[
\begin{cases}
x^3 + y^3 = 19 \\
x^2 + y^2 + 5x + 5y + xy = 12
\end{cases}
\]
we will use the substitution \( x + y = a \) and \( xy = b \).

1. **Express the first equation in terms of \(a\) and \(b\):**
   \[
   x^3 + y^3 = (x+y)(x^2 - xy + y^2) = a(a^2 - 3b)
   \]
   Given \( x^3 + y^3 = 19 \), we have:
   \[
   a(a^2 - 3b) = 19
   \]

2. **Express the second equation in terms of \(a\) and \(b\):**
   \[
   x^2 + y^2 + 5x + 5y + xy = 12
   \]
   We know that:
   \[
   x^2 + y^2 = (x+y)^2 - 2xy = a^2 - 2b
   \]
   Substituting this into the second equation:
   \[
   a^2 - 2b + 5a + b = 12 \implies a^2 - b + 5a = 12
   \]

3. **Solve for \(b\) in terms of \(a\):**
   \[
   a^2 - b + 5a = 12 \implies b = a^2 + 5a - 12
   \]

4. **Substitute \(b\) into the first equation:**
   \[
   a(a^2 - 3b) = 19 \implies a(a^2 - 3(a^2 + 5a - 12)) = 19
   \]
   Simplify the equation:
   \[
   a(a^2 - 3a^2 - 15a + 36) = 19 \implies a(-2a^2 - 15a + 36) = 19
   \]

5. **Solve the resulting polynomial equation:**
   \[
   -2a^3 - 15a^2 + 36a - 19 = 0
   \]
   We need to find the roots of this cubic equation. By inspection or using the Rational Root Theorem, we find that \(a = 1\) is a root.

6. **Verify \(a = 1\):**
   \[
   b = 1^2 + 5(1) - 12 = 1 + 5 - 12 = -6
   \]
   So, \(x\) and \(y\) satisfy:
   \[
   x + y = 1 \quad \text{and} \quad xy = -6
   \]
   The quadratic equation is:
   \[
   t^2 - (x+y)t + xy = t^2 - t - 6 = 0
   \]
   Solving for \(t\):
   \[
   t = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}
   \]
   Thus, \(t = 3\) or \(t = -2\). Therefore, the solutions are:
   \[
   (x, y) = (3, -2) \quad \text{or} \quad (-2, 3)
   \]

7. **Check for other potential roots:**
   The other potential root \(a = -\frac{19}{2}\) leads to:
   \[
   b = \left(-\frac{19}{2}\right)^2 + 5\left(-\frac{19}{2}\right) - 12 = \frac{361}{4} - \frac{95}{2} - 12 = \frac{361 - 190 - 48}{4} = \frac{123}{4}
   \]
   However, for real roots \(x\) and \(y\), we need:
   \[
   (x+y)^2 \geq 4xy \implies \left(-\frac{19}{2}\right)^2 \geq 4 \cdot \frac{123}{4} \implies \frac{361}{4} \geq 123
   \]
   This inequality holds, but solving the quadratic equation for these values does not yield real roots.

The final answer is \(\boxed{(3, -2) \text{ or } (-2, 3)}\)