[Siren-Of-Step]

2. Gives $\sqrt{2x-1} =t$
We have $\sqrt{t^2-6t+9} + \sqrt{t^2-2t+1}$
then$ \left|\,\right. t-3 \left|\,\right. +\left|\,\right. t-1\left|\,\right. =4$
Cases 1. $t < 1 , -t+1-t+3 = 4 , -2t = 0 , t = 0$
2. $1 \leqslant t \leqslant 3, t-1-t+3 = 4 , 2=4$
3. $t>3 , t-1+t-3 = 4 , t=4$

Union all cases , $t=0,4$ ....
** I Haven't checked the answers yet