SOLUTION: find three consecutive even integers such that the square of the third is 28 more than the square of the second.

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Question 311687: find three consecutive even integers such that the square of the third is 28 more than the square of the second.
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
let the numbers be x, x+2, x+4
square of the third= (x+4)^2
square of the second = (x+2)^2
.
(x+4)2 = (x+2)^2 +28
x^2 +8x +64= X^2+4x+4 +28
8x-4x=-64+32
4x= -32
x=-8
-8. -6, -4