1. **State the problem:** Solve for $v$ in the equation $$\sqrt{-3v + 40} = v - 4.$$\n\n2. **Understand the domain:** The expression inside the square root must be non-negative, so $$-3v + 40 \geq 0 \implies v \leq \frac{40}{3}.$$ Also, since the right side is $v - 4$, and the left side is a square root (always non-negative), we must have $$v - 4 \geq 0 \implies v \geq 4.$$\n\n3. **Combine domain restrictions:** $$4 \leq v \leq \frac{40}{3}.$$\n\n4. **Square both sides to eliminate the square root:**\n$$\left(\sqrt{-3v + 40}\right)^2 = (v - 4)^2$$\n$$-3v + 40 = (v - 4)^2$$\n\n5. **Expand the right side:**\n$$(v - 4)^2 = v^2 - 8v + 16$$\n\n6. **Rewrite the equation:**\n$$-3v + 40 = v^2 - 8v + 16$$\n\n7. **Bring all terms to one side:**\n$$0 = v^2 - 8v + 16 + 3v - 40$$\n$$0 = v^2 - 5v - 24$$\n\n8. **Solve the quadratic equation:**\nUse the quadratic formula $$v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-5$, $c=-24$.\n\nCalculate the discriminant:\n$$\Delta = (-5)^2 - 4(1)(-24) = 25 + 96 = 121$$\n\nCalculate the roots:\n$$v = \frac{5 \pm \sqrt{121}}{2} = \frac{5 \pm 11}{2}$$\n\nSo,\n$$v_1 = \frac{5 + 11}{2} = 8$$\n$$v_2 = \frac{5 - 11}{2} = -3$$\n\n9. **Check solutions against domain and original equation:**\n- For $v=8$: Check domain $4 \leq 8 \leq \frac{40}{3}$ (true). Check original equation:\n$$\sqrt{-3(8) + 40} = \sqrt{-24 + 40} = \sqrt{16} = 4$$\n$$v - 4 = 8 - 4 = 4$$\nBoth sides equal 4, so $v=8$ is a valid solution.\n\n- For $v=-3$: Check domain $4 \leq -3 \leq \frac{40}{3}$ (false), so discard $v=-3$.\n\n**Final answer:** $$v = 8.$$