Question 286359
Let x= the first integer. 
    x+1= the second integer, because they are consecutive integers

x(x+1) = (x-3)(x+1+4)
{{{x^2+x = (x-3)(x+5)}}}
{{{x^2+x = x^2 + 2x - 15}}}
{{{x^2-x^2 = 2x - x - 15}}}
{{{0=x-15}}}
{{{x=15}}}

so the equation looks like this:

x(x+1)=(x-3)(x+5)
(15)(16) = (12)(20)

The four numbers are 15,16,12, and 20. Their sum is 63.