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

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