<PRE><B>Question 48776</B>
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 the 3 CONSECUTIVE EVEN  integers be 2n,2n+2,2n+4
SQUARE OF THIRD = (2N+4)^2
SQUARE OF SECOND=(2N+2)^2
DIFFERENCE =(2N+4)^2-(2N+2)^2=76=4(N+2)^2-4(N+1)^2
(N+2)^2-(N+1)^2=19
(N+2+N+1)(N+2-N-1)=19
2N+3=19
2N=16
N=8
HENCE THE NUMBERS ARE 
2*8=16,18,20</PRE>