The sum of squares of solutions of equation \({x}^{2}-|2x-3|-4=0\). (28 Jan, Shift I, Memory Based)

\({x}^{2}-|2x-3|-4=0\\ |2x-3|=\left{\begin{matrix}2x-3,x\geq \frac{3}{2} \\ -2x+3,x<\frac{3}{2}\end{matrix}\right.\\ CaseI:whenx\geq \frac{3}{2}\\ {x}^{2}-2x-1=0\\ x=\frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2}\\ but1-\sqrt{2}isdiscarded\sin cexshouldbegreaterthan3/2.\\ So,x=1+\sqrt{2}\\ NowSimilarlyfromcaseII,wegetx=-1-2\sqrt{2}\\ NowSumofsquareofroots={{x}_{1}}^{2}+{{x}_{2}}^{2}\\ ={(1+\sqrt{2})}^{2}+{(-1-2\sqrt{2})}^{2}\\ =1+2+2\sqrt{2}+1+8+4\sqrt{2}\\ =12+6\sqrt{2}\\\)