Let us group the LHS in #(x-1)(2x+1)(x+1)(2x-3)=15# as follows
#{(x-1)(2x+1)}xx{(x+1)(2x-3)}=15#
or #(2x^2-2x+x-1)(2x^2+2x-3x-3)=15#
or #(2x^2-x-1)(2x^2-x-3)=15#
Now let us assume #2x^2-x=t#, then above equation becomes
#(t-1)(t-3)=15# or #t^2-4t+3=15#
or #t^2-4t-12=0# or #(t-6)(t+2)=0#
Hence #t=6# or #t=-2#
If #t=6#, then #2x^2-x=6# or #2x^2-x-6=0#
Here discriminant is #(-1)^2-4*2*(-6)=49# which is square of a ratioonal number and we can get rational roots. The above equation becomes
#(2x+3)(x-2)=0# i.e. #x=2# or #x=-3/2#
If #t=-2#, then #2x^2-x=-2# or #2x^2-x+2=0#
Here as discriminant is #(-1)^2-4*2*2=-15# and as it is negative, we do not have real roots.
Hence, only real roots of #(x-1)(2x+1)(x+1)(2x-3)=15# are #{2,-3/2}#.
