Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$

Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$

I just know $x^{x^x}$ is increasing in $x$ and hence the equation has a unique solution, nut then I dont know how to move on, I also know viete' formula but I dont know if it helps here, thanks in advance.

The sum is $-2$. Can you see why? Hint: I have not computed the solutions.

Details: As you observed, there is a unique positive $b$ such that $b^{(b^b)}=2012$. Moreover, this $b$ is in the interval $(2,3)$, by the Intermediate Value Theorem, since $2^{(2^2)}$ is too small and $3^{(3^3)}$ is too big.

Note that $x^2+2x+3=(x+1)^2+2$, so $x^2+2x+3$ attains a minimum value of $2$. Thus the equation $x^2+2x+3-b=0$ has two real solutions. The sum of these is the negative of the coefficient of $x$, that is, $-2$.

Hint: Look at Vieta's Formula then compare your equation.