1.3x^2-2 root sign 5x+5 = 0 ,has no solution in the real number range and has a solution in the complex number range. Now solve by complex numbers:

3[(x-(root 5/3)]^2-5/3+5 = 0

(x-root 5/3)^2 = -10/9

x-root 5/3 = ±[(root 10)/3]i

Therefore, x = 1/3[root 5±(root 10)i] ,that is x = (root 5/3)(1+ root 2i)

2.2T^2-7T-4 = 0,

2(T-7/4)^2-49/8-4 = 0

(T-7/4)^2 = 81/16, T-7/4 = ±9/4, so T=±9/4+7/4

Therefore, T1=4, T2=-1/2

3.3(x-2/3)^2+2/3 = 0 , same as question 1, solved by complex numbers:

(x-2/3)^2 = -2/9,x = 1/3[2±(root 2)i]

X2-5X+6 = 0

x^2-5x+25/4=25/4-6

(x-5/2)^2=1/4

x-5/2=1/2 or -1/2

x=3 or 2

1):

y^+4 root 3y-2=0

2):3x^-4x-2=0

3):(x-1)^-6(x-1)+8=0

< p>1)=(y+2 root 3)^-10=0

then y+2 root 3=positive and negative root 10

2)=3(x-2/3)^-10/3=0

then x-2/3=......

3)Let x-1=a,then a^-6a+8 =0 (a-3)^-1=0 a-3= plus or minus 1 a=4 or 2

x=7 or 5