We know that #x_1x_2x_3x_4=12#
now making #x_1x_2 = 2# then #x_3x_4=6#
Observing #3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 0# by inspection we find that #x = 6# is a root then making #x_1=6# we obtain #x_2 = 1/3# and a we can verify #x=1/3# is also a root. So we have
#3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 3(x-6)(x-1/3)(x^2+alpha x+beta)#
so grouping coefficients we have
#{(6 beta=12), (19 beta - 6 alpha= 50),
(19 alpha - 3 beta=-44), (3 alpha=-6):}#
and solving we have
#alpha=-2, beta=2# then
#3 x^4 - 25 x^3 + 50 x^2 - 50 x + 12 = 3(x-6)(x-1/3)(x^2-2 x+2)#
and
#x^2-2x+2=0# has two complex roots
#x = 1 pm i#