$$\begin{aligned} & \mathrm{x}^2+|2 \mathrm{x}-3|-4=0 \\ & \text { Case } \mathrm{I}: \mathrm{x} \geq \frac{3}{2} \\ & \qquad \begin{array}{l} \mathrm{x}^2+2 \mathrm{x}-3-4=0 \\ \mathrm{x}^2+2 \mathrm{x}-7=0 \\ \mathrm{x}=2 \sqrt{2}-1 \end{array} \end{aligned}$$

$$\begin{aligned} & \text { Case } \mathrm{II}: x<\frac{3}{2} \\ & \begin{aligned} & x^2+3-2 x-4=0 \\ & x^2-2 x-1=0 \\ & x=1-\sqrt{2} \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { Sum of squares } & =(2 \sqrt{2}-1)^2+(1-\sqrt{2})^2 \\ & =8-4 \sqrt{2}+1+1-2 \sqrt{2}+2 \\ & =6(2-\sqrt{2}) \quad \therefore(3) \end{aligned}$$