Find three consecutive positive integers such that the product of the first and second is 37 less thanthe square of the third

Consecutive integers are 1 apart

they are x, x+1, and x+2

produce of first and 2nd (x and x+1) is 37 less than square of third (x+2)

(x) (x+1) = - 37 + (x+2) ²

expand

x²+x=x²+4x+4-37

x²+x=x²+4x-33

minu x² from both sides

x=4x-33

minus x both sides

0=3x-33

add 33 both sides

33=3x

divide by 3 both sides

11=x

x+1=12

x+2=13

the numbers are 11,12,13

See the pic for the answer.