Question 112530
Because we need 3 consecutive numbers, I would call them x, x+1, x+2.  This way we will be solving for x, the smallest number.  So, using the clues about the numbers, we get the equation 2x^2 = (x+1)(x+2) + 38.
Simplifying gives 2x^2 = x^2 + 3x + 2 + 38
2x^2 = x^2 + 3x + 40
Now we solve this quadratic equation.  We could subtract the right side of the equation from both sides.
x^2 - 3x - 40 = 0
We could factor this to (x-8)(x+5)=0
So x=8 or x=-5.  You just need 1 set of numbers, so I would say 8, 9, 10.
You could check this: 2*(8)^2=128 and 9*10+38 = 128, so this set of numbers works.