Question 148763
Three consecutive even integers are such that the square of the third is 76 more than the square of the second. Find the three integers. 
.
Let x = 1st consecutive even integer
x+2 = 2nd consecutive even integer
x+4 = 3rd consecutive even integer
.
(x+4)^2 - 76 = (x+2)^2
expanding the polynomial:
x^2+8x+16 - 76 = x^2+4x+4
subtracting x^2 from both sides:
8x+16 - 76 = 4x+4
8x - 60 = 4x+4
8x = 4x+64
4x = 64
x = 16
.
Therefore the three even integers are:
16, 18, and 20