Given, equation x2−2(1+3m)x+7(3+2m)=0 Here, a>0 and D=b2−4ac>0 Expression is always positive, if roots are distinct then D>0, ‌[−2(1+3m)]2−4×7(3+2m)>0 ⇒4[1+9m2+6m]−84−56m>0 ⇒36m2−32m−80>0 ⇒9m2−8m−20>0 ⇒(m−2)(m+‌
10
9
)>0 ⇒m∈(−∞,‌
−10
9
)∪[2,∞) ∴ Integral value of m are .....3i,1,2,3,4,5,6. So, the number of element in S is infinite.