Question 253383
Let the three consecutive even integers be x,x+2 and x+4
Given, the square of the first number decreased by 3 times the second plus the third number is 22
          x^2-3(x+2)+(x+4) = 22
          x^2-3x-6+x+4 = 22
          x^2-2x-24 = 0
          (x-6)(x+4) = 0
                   x = 6 or -4  
x cannot be negative.
Therefore x = 6.
So the given numbers are 6,8,10