Find three consecutive odd integers such that three times the square of the first integer is twelve more than the product of the second and the third integers

Consecutive odd integers are 2 apart
they are
n,n+2,n+4

3(n²)=12+(n+2)(n+4)
expand
3n²=12+n²+6n+8
3n²=n²+6n+20
minus n² both sides
2n²=6n+20
divide bot sides by 2
n²=3n+10
minus 3n+10 both sides
n²-3n-10=0
factor
(n-5)(n+2)=0
n-5=0
n=5

n+2=0
n=-2
has to be odd, so no

n=5
n+2=7
n+4=9

the integers are 5,7,9

I’m not sure i’m sorry