Given x = 2

Putting this value in each of the options, we check where LHS = RHS

A. x + 2 = 5

LHS = x +2 = 2 +2 =4

RHS = 5

LHS ≠RHS

B. x - 2 = 0

LHS = x - 2 = 2 - 2 = 0

RHS = 0

LHS = RHS

C. 2x + 1 = 0

LHS = 2x +1 = 2× 2 + 1=4 + 1 =5

RHS = 0

LHS ≠RHS

D. x + 3 = 6

LHS = x +3 = 2 +3 = 5

RHS = 6

LHS ≠RHS

Hence, x =2 is the solution of x-2 =0.