Four consecutive integers... n, n+1, n+2, n+3.
The product of the 1st and 4th is four less than twice the 1st multiplied by the 4th.
n(n+3)=2n(n+3)-4 perform indicated multiplications...
n^2+3n=2n^2+6n-4 subtract n^2 from both sides
3n=n^2+6n-4 subtract 3n from both sides
n^2+3n-4=0 factor
n^2-n+4n-4=0
n(n-1)+4(n-1)=0
(n+4)(n-1)=0, and since n>0
n=1
So the four numbers are 1, 2, 3, 4
check...
1(4)=2(1)4-4
4=8-4
4=4
