quadratic absolute value

Observe that for all real x, the solution must strictly be $x<-5/2$ otherwise, the equation fails.

Here , we explore 2 cases:

Case 1: $|x^{2}+4x+3|=-(2x+5)|$

Case 2: $x^{2}+4x+3=(2x+5)$

Taking case 1 for instance:

$|x^{2}+4x+3|=-(2x+5)|$ $x^{2}+6x+8=0$ which gives roots $-4 ,-2$ but $x<-5/2$ so, only $-4$ is accepted

Taking case 2 , we have:

$x^{2}+2x-2=0$ Which of course has roots $-1+\sqrt{3}$ and $-1-\sqrt{3}$ only $-1-\sqrt{3}$ qualifies because again x must satisfy $x<-5/2$

Hence the equation has roots -4 and $x<-5/2$

Fulfilling the question, we have: $-4+3-1=-2$