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

See the pic for the answer.

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