To solve the compound inequality \(6c + 8 \leq -10\) or \(8c + 12 > 20\), we will solve each inequality separately.
1. Solving the first inequality:
\[ 6c + 8 \leq -10 \]
Subtract 8 from both sides:
\[ 6c \leq -10 - 8 \]
\[ 6c \leq -18 \]
Now, divide both sides by 6:
\[ c \leq -3 \]
2. Solving the second inequality:
\[ 8c + 12 > 20 \]
Subtract 12 from both sides:
\[ 8c > 20 - 12 \]
\[ 8c > 8 \]
Now, divide both sides by 8:
\[ c > 1 \]
3. Combining the solutions:
We have two separate solutions:
- From the first inequality, \(c \leq -3\)
- From the second inequality, \(c > 1\)
Since it is an "or" statement, the final solution is:
\[ c \leq -3 \quad \text{or} \quad c > 1 \]
Thus, the solution to the compound inequality is:
\[ \boxed{c \leq -3 \text{ or } c > 1} \]