Пусть (t²-2t)=х, а (t²-2t)²= х²
х²-3=2х
х²-2х-3=0
D=4-4*(-3)=4+12=16
х1= \frac{2+4}{2} = \frac{6}{2} =3
х2= \frac{2-4}{2} = \frac{-2}{2} =-1
(t²-2t)=3
t²-2t-3=0
D=4D=4-4*(-3)=4+12=16
х1= \frac{2+4}{2} = \frac{6}{2} =3
х2= \frac{2-4}{2} = \frac{-2}{2} =-1
(t²-2t)=-1
t²-2t+1=0
D=4-4=0
х1,2= \frac{2}{2} =1