Square Root of Irrational Expression

We are given $(a + b\sqrt{n})^2 = 29 - 12\sqrt{5}$. $a^2 + b^2 n + 2ab\sqrt{n} = 29 - 12\sqrt{5}$ So, $a^2 + b^2 n = 29$ and $2ab\sqrt{n} = -12\sqrt{5} = -2\sqrt{180}$.

Case 1: $n = 5 \implies ab = -6$. If $b = 2$, $a^2 + 5(4) = 29 \implies a^2 = 9 \implies a = -3$ (since $a + b\sqrt{n} > 0$). Here, $(a, b, n) = (-3, 2, 5)$, so $a + b + n = -3 + 2 + 5 = 4$.

Case 2: $n = 20 \implies 2ab\sqrt{5} = -6\sqrt{5} \implies ab = -3$. If $b = 1$, $a^2 + 20(1) = 29 \implies a^2 = 9 \implies a = -3$. Here, $(a, b, n) = (-3, 1, 20)$, so $a + b + n = -3 + 1 + 20 = 18$.

Hence, the maximum possible value of $(a + b + n)$ is $18$.