The tutor above interpreted the problem wrong.  He interpreted it as if it had read:

But it's this:

Instead of doing yours for you, I'll do this one instead which is
done exactly the same:

x = the smallest
x+2 = next to smallest
x+4 = next to largest
x+6 = largest

 
So we add them all up by putting in plusses 

x + (x+2) + (x+4) + (x+6)

 
That means to put an equal sign after it.
So we have:

x + (x+2) + (x+4) + (x+6) =

 
So we write 3∙ after the equal sign, and we have:

x + (x+2) + (x+4) + (x+6) = 3∙

 
That means to subtract them in that order.  Since the next to the smallest is
(x+2) is mentioned first and 20 second, we subtract them in that order and put
[(x+2)-20] after the 3∙

So the equation is

x + (x+2) + (x+4) + (x+6) = 3∙[(x+2)-20]

x + x + 2 + x + 4 + x + 6 = 3∙[x + 2 - 20]

4x + 12 = 3∙[x - 18]

4x + 12 = 3x - 54

      x = -66

  x = the smallest = -66
x+2 = next to smallest = -66+2 = -64
x+4 = next to largest = -66+4 = -62
x+6 = largest = -66+6 = -60

Now do yours, which has only 3 consecutive numbers.

[Your answers will be negative too]

Edwin

The answer the other person dkppathak(420) gave you makes ABSOLUTELY no SENSE!!

Let the first integer be F
Then other 2 are: F + 2, and F + 4
As these numbers are INTEGERS, the CORRECT equation should NOT be: F + F + 2 + F + 4 = 2(14 - F)

Instead, it should be: F + F + 2 + F + 4 = 2(F - 14)
3F + 6 = 2F - 28
3F - 2F = - 28 - 6
First integer, or 

Now, it shouldn't be too difficult to find the other 2 integers!