(x^2-3x)^2-(x^2-3x)-8=0
x^2-3x=t
t^2-t-8=0
D=1+4*8=33
t_1=(1+sqrt(33))/2
t_2=(1-sqrt(33))/2
x^2-3x-(1+sqrt(33))/2=0
D=9+2+2sqrt(33)=11+2sqrt(33)
x_12=(3+-sqrt(11+2sqrt(33))/2
x^2-3x+(sqrt(33)-1)/2=0
D=9-2sqrt(33)+2<0
ответ: (3+-sqrt(11+2sqrt(33))/2
