1. **State the problem:** Solve for all possible values of $x$ in the equation $$\sqrt{6x + 40} = x + 8.$$\n\n2. **Recall the domain and formula:** The square root function $\sqrt{y}$ is defined only for $y \geq 0$, so we require $6x + 40 \geq 0$. Also, since the right side is $x + 8$, it must be non-negative because the square root is always non-negative. Thus, $x + 8 \geq 0$.\n\n3. **Write down the domain restrictions:**\n- $6x + 40 \geq 0 \implies x \geq -\frac{40}{6} = -\frac{20}{3} \approx -6.67$.\n- $x + 8 \geq 0 \implies x \geq -8$.\n\nThe combined domain is $x \geq -\frac{20}{3}$ since $-\frac{20}{3} > -8$.\n\n4. **Square both sides to eliminate the square root:**\n$$\left(\sqrt{6x + 40}\right)^2 = (x + 8)^2$$\n$$6x + 40 = (x + 8)^2$$\n\n5. **Expand the right side:**\n$$6x + 40 = x^2 + 16x + 64$$\n\n6. **Bring all terms to one side to form a quadratic equation:**\n$$0 = x^2 + 16x + 64 - 6x - 40$$\n$$0 = x^2 + 10x + 24$$\n\n7. **Factor the quadratic:**\n$$x^2 + 10x + 24 = (x + 6)(x + 4)$$\n\n8. **Set each factor equal to zero:**\n$$x + 6 = 0 \implies x = -6$$\n$$x + 4 = 0 \implies x = -4$$\n\n9. **Check each solution against the original equation and domain:**\n- For $x = -6$:\n$$\sqrt{6(-6) + 40} = \sqrt{-36 + 40} = \sqrt{4} = 2$$\n$$x + 8 = -6 + 8 = 2$$\nBoth sides equal 2, so $x = -6$ is valid.\n\n- For $x = -4$:\n$$\sqrt{6(-4) + 40} = \sqrt{-24 + 40} = \sqrt{16} = 4$$\n$$x + 8 = -4 + 8 = 4$$\nBoth sides equal 4, so $x = -4$ is valid.\n\n**Final answer:** $$x = -6 \text{ or } x = -4.$$