$$
\begin{aligned}
&x^2+|2 x-3|-4=0 \\
&|2 x-3|=\left\{\begin{array}{lll}
(2 x-3) & \text { if } & \mathrm{x}>\frac{3}{2} \\
-(2 x-3) & \text { if } & \mathrm{x} < \frac{3}{2}
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
&\text { for } x>\frac{3}{2}, \quad x^2+2 x-3-4=0 \\
&x^2+2 x-7=0 \\
&x=\frac{-2 \pm \sqrt{4+28}}{2} \\
&=\frac{-2 \pm 4 \sqrt{2}}{2}=-1 \pm 2 \sqrt{2} \\
&\text { Here } x=2 \sqrt{2}-1 \\
&\left\{2 \sqrt{2}-1 < \frac{3}{2}\right\} \\
&\text { for } x < \frac{3}{2} \\
&x^2-2 x+3-4=0 \\
&\Rightarrow x^2-2 x-1=0 \\
&\Rightarrow x=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2} \\
&\text { Here } x=1-\sqrt{2} \quad\left\{(1-\sqrt{2}) < \frac{3}{2}\right\}
\end{aligned}
$$
Sum of roots: $(2 \sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}$