Find four consecutive integers such that the product of the two largest is 46 more than the product of the two smallest integers

Let x, x+1, x+2 & x+3 be the four integers.

(x+2) (x+3) = x (x+1) + 46

x^2+5x+6=x^2+x+46

x^2-x^2+5x-x=46-6

4x=40

x=40/4

x=10 for the smallest integer.

10=1=11 for the next integer.

10=2=12 for the third integer.

10=3=13 for the largest integer.

12*13=10*11+46

156=110+46

156=156