## Elementary Algebra

We square both sides of the equation and solve. Thus: $\sqrt{2n+1} = 2 + \sqrt{n-3} \\ 2n+1 = 4 + n -3 + 4\sqrt{n-3} \\ n = 4 \sqrt{n-3} \\ n^2 = 16(n-3) \\ n^2 - 16n + 48 = 0 \\ (n-4)(n-12) \\ n = 4, 12$