Find two consecutive integers whose product is 24 more than three times the first integer

X= first integer
x+1= second consecutive integer

Product means multiply, so we'll multiply x by x+1. More than means to add and 3 times is multiplication.

x(x + 1)= 3x + 24
use distributive property to multiply x by everything in parentheses
(x*x) + (x*1)= 3x + 24
x^2 + x= 3x + 24
subtract 3x from both sides
x^2 - 2x= 24
subtract 24 from both sides
x^2 - 2x - 24= 0
factor the expression
(x-6)(x+4)= 0
x-6=0
x=6

x+4=0
x=-4

If x= 6 is the first integer, x+1= 7 is the second integer

If x= -4 is the first integer, x+1= -3 is the second integer

Hope this helps! :)

4should be the last one n=4