How many valid solutions does the equation √(x+10) = x - 2 have, after rejecting any extraneous candidates?

Square both sides of $\sqrt{x + 10} = x - 2$: $x + 10 = (x - 2)^2 = x^2 - 4x + 4$. Rearranging gives $x^2 - 5x - 6 = 0$, so $(x - 6)(x + 1) = 0$ and the candidates are $x = 6$ and $x = -1$. Check each in the original. For $x = 6$: $\sqrt{16} = 4$ and $6 - 2 = 4$, so it works. For $x = -1$: $\sqrt{9} = 3$ but $-1 - 2 = -3$, so it fails. Only one valid solution remains, choice D. Choice A counts the extraneous root. Choice B is wrong since $x = 6$ works. Choice C is impossible for a radical equation.