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
