Find all the integers $x$ in $[20, 50]$ such that $6x+5\equiv 19 \mod 10$, that is, $10$ divides $(6x+15)+19$.
1. We start with the given congruence:
   \[
   6x + 5 \equiv 19 \pmod{10}
   \]
2. Subtract 5 from both sides to simplify:
   \[
   6x \equiv 14 \pmod{10}
   \]
3. Reduce \(14 \pmod{10}\):
   \[
   14 \equiv 4 \pmod{10}
   \]
   So, we have:
   \[
   6x \equiv 4 \pmod{10}
   \]
4. To simplify further, we divide both sides by 2 (since 2 is a common factor of 6 and 4):
   \[
   3x \equiv 2 \pmod{5}
   \]
5. Next, we need to find the multiplicative inverse of 3 modulo 5. We need a number \(y\) such that:
   \[
   3y \equiv 1 \pmod{5}
   \]
   By testing, we find that:
   \[
   3 \cdot 2 = 6 \equiv 1 \pmod{5}
   \]
   So, the inverse of 3 modulo 5 is 2.
6. Multiply both sides of \(3x \equiv 2 \pmod{5}\) by 2:
   \[
   2 \cdot 3x \equiv 2 \cdot 2 \pmod{5}
   \]
   Simplifying, we get:
   \[
   6x \equiv 4 \pmod{5}
   \]
   Since \(6 \equiv 1 \pmod{5}\), we have:
   \[
   x \equiv 4 \pmod{5}
   \]
7. Therefore, \(x\) can be written as:
   \[
   x = 5k + 4 \quad \text{for some integer } k
   \]
8. We need to find all integers \(x\) in the interval \([20, 50]\). We substitute \(x = 5k + 4\) and solve for \(k\):
   \[
   20 \leq 5k + 4 \leq 50
   \]
   Subtract 4 from all parts:
   \[
   16 \leq 5k \leq 46
   \]
   Divide by 5:
   \[
   3.2 \leq k \leq 9.2
   \]
   Since \(k\) must be an integer, we have:
   \[
   k = 4, 5, 6, 7, 8, 9
   \]
9. Substitute these values of \(k\) back into \(x = 5k + 4\):
   \[
   \begin{align*}
   k = 4 & \implies x = 5 \cdot 4 + 4 = 24 \\
   k = 5 & \implies x = 5 \cdot 5 + 4 = 29 \\
   k = 6 & \implies x = 5 \cdot 6 + 4 = 34 \\
   k = 7 & \implies x = 5 \cdot 7 + 4 = 39 \\
   k = 8 & \implies x = 5 \cdot 8 + 4 = 44 \\
   k = 9 & \implies x = 5 \cdot 9 + 4 = 49 \\
   \end{align*}
   \]

The final answer is \(\boxed{24, 29, 34, 39, 44, 49}\).