1. We are given the inequality $6 \leq -3(4-2x) < 8$ and need to solve for $x$. 2. Start by distributing $-3$ inside the parenthesis: $$-3(4-2x) = -3 \times 4 + (-3) \times (-2x) = -12 + 6x$$ 3. Substitute back into the inequality: $$6 \leq -12 + 6x < 8$$ 4. Now, add $12$ to all parts of the inequality to isolate the term with $x$: $$6 + 12 \leq -12 + 6x + 12 < 8 + 12$$ $$18 \leq 6x < 20$$ 5. Divide all parts by $6$ to solve for $x$: $$\frac{18}{6} \leq \frac{6x}{6} < \frac{20}{6}$$ $$3 \leq x < \frac{10}{3}$$ 6. Therefore, the solution to the inequality is: $$x \in [3, \frac{10}{3})$$