Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers?

((x+6)(x+8))-((X + 2)(X+4)) = 72

(x^2 + 14x + 48)-(x^2 + 6x + 8) = 72

8x + 40 = 72

8x = 32

x = 4

check:

((10)(12))-((6)(8)) = 72

(120)-(48) = 72

120 = 120

12,10,8,6
The numbers are 6, 8, 10, and 12.

Here is how we find them:

(n + 6)(n + 4) - (n + 2)n = 72; whence,

(n2 + 10n + 24) - (n2 + 2n) = 8n + 24 = 72,

8n = 48, and

n = 6.

So, let's try it!

(12)(10) - (8)(6) = 120 - 48 = 72.

It works!