1. **State the problem:** We have an equilateral triangle with vertices\nA(0, -2), B(0, -10), and C(4\sqrt{3}, y), where $y < 0$. We need to find the value of $y$.\n\n2. **Calculate side lengths:** Since the triangle is equilateral, all sides are equal in length.\nFirst, find length AB:\n$$AB = \sqrt{(0-0)^2 + (-10+2)^2} = \sqrt{0 + (-8)^2} = 8.$$\n\n3. **Express lengths AC and BC in terms of $y$:**\n$$AC = \sqrt{(4\sqrt{3} - 0)^2 + (y + 2)^2} = \sqrt{(4\sqrt{3})^2 + (y + 2)^2} = \sqrt{48 + (y+2)^2}.$$\n$$BC = \sqrt{(4\sqrt{3} - 0)^2 + (y + 10)^2} = \sqrt{48 + (y + 10)^2}.$$\n\n4. **Set side lengths equal:** Since $AB = AC = BC = 8$, set the squares equal to $8^2 = 64$ to avoid square roots:\n$$48 + (y+2)^2 = 64,$$\n$$48 + (y+10)^2 = 64.$$\n\n5. **Solve the first equation:**\n$$48 + (y+2)^2 = 64 \implies (y+2)^2 = 16 \implies y+2 = \pm 4.$$\nSo,\n$y = -2 \pm 4$. Possible values: $y = 2$ or $y = -6$.\n\n6. **Solve the second equation:**\n$$48 + (y+10)^2 = 64 \implies (y+10)^2 = 16 \implies y + 10 = \pm 4.$$\nSo,\n$y = -10 \pm 4$. Possible values: $y = -6$ or $y = -14$.\n\n7. **Find common $y$:** The only common value satisfying both equations is $y = -6$.\nAlso, given $y < 0$, this solution meets the condition.\n\n**Final answer:** $$\boxed{-6}.$$