S={x |x^2-3x+2=0},
={ x| (x-1)(x-2)=0 }
={ 1,2}
T={a|x2-2x-a=0,x∈R,a∈R}
if a=1
x2-2x-1=0
△ = 4+4 = 8 >0
real roots
=> 1 ∈ T
if a=2
x2-2x-2=0
△ = 4+ 8 = 12 >0
real roots
=> 2 ∈ T
1,2∈ T and S={1,2}
=> S∩ T = {1,2} #