Sum of the roots of the equation 

$x^2+|2x-3|-4=0$ is

Case 1: When $x<3/2$ 

In this case the equation becomes 

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2-(2x-3)-4=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2-2x+1=2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(x-1{)}^2=2\Rightarrow x-1=\pm \sqrt{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=1\pm \sqrt{2}\end{array}$

As $x<3/2$ we take, $x=1-\sqrt{2}$

Case 2: When $x\ge 3/2$ 

In this case the equation becomes

$\begin{array}{l}{x}^2+(2x-3)-4=0\\ \Rightarrow (x+1{)}^2=8\Rightarrow x+1\pm 2\sqrt{2}\end{array}$

As $x\ge 3/2,x=-1+2\sqrt{2}$

Sum of the roots is 

                             $(1-\sqrt{2})+(-1+2\sqrt{2})=\sqrt{2}$.