(1)移项得:(x-1)2=4,
开方得:x-1=±2,
解得:x1=3,x2=-1;
(2)2x2-5x-1=0,
∵b2-4ac=(-5)2-4×2×(-1)=33,
∴x=
5±
33
2×2,
即x1=
5+
33
4,x2=
5−
33
4.
(1)移项得:(x-1)2=4,
开方得:x-1=±2,
解得:x1=3,x2=-1;
(2)2x2-5x-1=0,
∵b2-4ac=(-5)2-4×2×(-1)=33,
∴x=
5±
33
2×2,
即x1=
5+
33
4,x2=
5−
33
4.