Find four consecutive positive integers such that the product of the first and fourth is four less than twice the first multiplied by the fourth.

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

D. i think it is the savings