In the problem, it's clear that it is the four times repeated composition of the function $f(x) = (x+2)^2$ and that $15129$ is the square of $123$. So, as we work backward, we eventually go from $15129$ to $123$ to $11$ to $3$ to $1$ and every time (except the last time), we neglect the corresponding negative answer because there is a perfect square on the LHS. So, finally, the answer should be $+1$ and $-1$. However, upon plugging in $(+1)$ to verify the solution, it turns out to be extraneous. Why does this happen? Where do we gain this extra solution?
Also, is there a better way to solve this problem using Vieta's formula, which may be more generally applicable to this variety of problems where, there exists a polynomial that can be manipulated into the repeated composition of a certain quadratic, and hence its roots can be analyzed easily?
Suppose $(x+2)^2= t$, $(2+t)^2 = t_2$, $(2+t_2)^2=t_3$ etc.
Then you're given that $(2+t_3)^2=123^2$ and it follows that $$(t_3+125)(t_3-121) =0$$ Then put the value $t_3$ and proceed: $$((2+t_2)^2+125)((2+t_2)^2-11^2)=0$$ The first term gives a complex solution (since it means $(2+t_2)^2=-125$), so it is not necessary for us to consider it.
So if we proceed again by factoring using difference of squares we get $$(t_2+13)(t_2-9)=0$$ Ignore the first term again and get: $$(t+5)(t-1) =0$$ Ignore the complex solution, so that $t-1$ factors into $$(x+3)(x+1)=0$$
These are the only real solutions. Then their sum is?