Find all the integers $x$ in $[20, 50]$ such that $6x + 5 \equiv -19 \mod 10,$ that is, $10$ divides $(6x + 15) + 19.$
1. Start with the given congruence equation:
   \[
   6x + 5 \equiv -19 \pmod{10}
   \]

2. Simplify the right-hand side of the congruence:
   \[
   -19 \equiv 1 \pmod{10}
   \]
   This is because \(-19 + 20 = 1\).

3. Substitute the simplified right-hand side back into the original congruence:
   \[
   6x + 5 \equiv 1 \pmod{10}
   \]

4. Subtract 5 from both sides to isolate the term involving \(x\):
   \[
   6x \equiv 1 - 5 \pmod{10}
   \]
   \[
   6x \equiv -4 \pmod{10}
   \]

5. Simplify \(-4\) modulo 10:
   \[
   -4 \equiv 6 \pmod{10}
   \]
   This is because \(-4 + 10 = 6\).

6. Substitute the simplified value back into the congruence:
   \[
   6x \equiv 6 \pmod{10}
   \]

7. Divide both sides by 6. Since 6 and 10 are not coprime, we need to check if the division is valid. Here, we can see that:
   \[
   6x \equiv 6 \pmod{10} \implies x \equiv 1 \pmod{5}
   \]
   This is because dividing both sides by 6 (which is equivalent to multiplying by the modular inverse of 6 modulo 10, which is 1) simplifies the equation.

8. Determine the values of \(x\) in the interval \([20, 50]\) that satisfy \(x \equiv 1 \pmod{5}\):
   \[
   x = 5k + 1 \quad \text{for integers } k
   \]
   We need \(20 \leq 5k + 1 \leq 50\).

9. Solve the inequalities:
   \[
   20 \leq 5k + 1 \implies 19 \leq 5k \implies k \geq 4
   \]
   \[
   5k + 1 \leq 50 \implies 5k \leq 49 \implies k \leq 9
   \]

10. Therefore, \(k\) ranges from 4 to 9. The corresponding values of \(x\) are:
    \[
    x = 5k + 1 \quad \text{for } k = 4, 5, 6, 7, 8, 9
    \]
    \[
    x = 5(4) + 1 = 21
    \]
    \[
    x = 5(5) + 1 = 26
    \]
    \[
    x = 5(6) + 1 = 31
    \]
    \[
    x = 5(7) + 1 = 36
    \]
    \[
    x = 5(8) + 1 = 41
    \]
    \[
    x = 5(9) + 1 = 46
    \]

Conclusion:
The integers \(x\) in the interval \([20, 50]\) that satisfy the given congruence are:
\[
\boxed{\{21, 26, 31, 36, 41, 46\}}
\]