System of equations

From the given system, we can conclude that the values of $x,$ $y$ and $z$ must be all greater than or equal to zero. Without losing generality we can assume that $x\leq y\leq z \quad (*)$ Therefore, $x^2\leq y^2\leq z^2.$ Then $2y+3\leq 2z+3 \leq 2x+3,$ and this implies that $y\leq z \leq x \quad (**)$ From the inequalities $(*)$ and $(**)$ , it is easy to see that $x=z.$ The latter equation together with $(*)$ , implies that $x=y= z.$ So any solution of the given system must be formed by a triple of the form $(t, t,t).$ To find $t$ we should solve the equation $t=\sqrt{2t+3},$ whose only solution is $t=3.$ Therefore the only possible solution of the given system is $(3, 3, 3).$ Then the answer for this question is $3+3+3=\boxed{9}.$

As the system of equations is symmetrical and complex square root expressions are involved, I assumed that $x = y = z$ .

Solving it we get the quadratic $x^2 - 2x -3 = 0$ which on solving yields $x = -1$ or $3$ .

Note that $x, y$ and $z$ are real so $x=y=z=-1$ is not a solution.

Thus the answer is $x = y = z = 3$ .