1. **State the problem:** Solve for all possible values of $x$ in the equation $$\sqrt{48 - 6x} = x - 8.$$\n\n2. **Understand the domain:** The expression under the square root must be non-negative, so $$48 - 6x \geq 0 \implies x \leq 8.$$ Also, since the right side is $x - 8$, and the square root is always non-negative, we must have $$x - 8 \geq 0 \implies x \geq 8.$$ Combining these, the only possible $x$ is $8$.\n\n3. **Solve the equation:** Square both sides to eliminate the square root: $$\left(\sqrt{48 - 6x}\right)^2 = (x - 8)^2 \implies 48 - 6x = (x - 8)^2.$$\n\n4. **Expand and simplify:** $$48 - 6x = x^2 - 16x + 64.$$ Rearranged: $$0 = x^2 - 16x + 64 + 6x - 48 = x^2 - 10x + 16.$$\n\n5. **Solve the quadratic:** Use the quadratic formula $$x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2} = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}.$$\n\n6. **Find roots:** $$x = \frac{10 + 6}{2} = 8$$ or $$x = \frac{10 - 6}{2} = 2.$$\n\n7. **Check solutions against domain:** From step 2, $x$ must be 8. Check $x=2$: $$\sqrt{48 - 6(2)} = \sqrt{48 - 12} = \sqrt{36} = 6,$$ but right side is $$2 - 8 = -6,$$ which is not equal. So $x=2$ is extraneous.\n\n8. **Final answer:** $$\boxed{8}.$$