The sum of four consecutive even integers is 44. what are the numbers? , pls

Let's go... They are consecutive and even integers.
Let call the first number as y.
y is an even integer. So its consecutive even integers must be:
y + 2, y + 4, y + 6. 
First even integer is y.
Second even integer is y + 2.
Third even integer is y + 4.
Fourth even integer is y + 6.

So, the sum is y + (y+2) + (y+4) + (y+6). This is equal to 44.
So, we have the equation:

y + (y+2) + (y+4) + (y+6) = 44

Let's add the terms...
y + y + y + y + 2 + 4 + 6 = 44
4y + 12 = 44

Let's subtract 12 from both sides...
4y + 12 - 12 = 44 - 12
4y = 32

Now, let's divide by 4...
4y/4 = 32/4

So, y = 8.

First even integer is y = 8
Second even integer is y + 2 => 8 + 2 = 10
Third even integer is y + 4 => 8 + 4 = 12
Fourth even integer is y + 6 => 8 + 6 = 14

So, the four consecutive even integers are 8, 10, 12 and 14.

The answer should be c.