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)+46x^2+5x+6=x^2+x+46x^2-x^2+5x-x=46-64x=40x=40/4x=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+46156=110+46156=156