Find 3 consecutive odd numbers where...?

x > y > z
x - 2 = y
y - 2 = z
yz = x^2 - 34

x - 2 = y
x = y + 2

yz = x^2 - 34
y(y - 2) = (y + 2)^2 - 34
y^2 - 2y = (y + 2)(y + 2) - 34
y^2 - 2y = y^2 + 2y + 2y + 4 - 34
y^2 - y^2 - 2y - 4y = 4 - 34
-6y = -30
y = -30/-6
y = 5

x - 2 = y
x - 2 = 5
x = 5 + 2
x = 7

y - 2 = z
5 - 2 = z
z = 3

∴ x = 7 , y = 5 , z = 3

let x=smallest number ; x+2=middle number ; x+4=largest number
x(x+2) = (x+4)^2 - 34
x^2 + 2x = x^2 + 8x + 16 - 34
isolate x in the right side
0 = x^2 - x^2 + 8x - 2x -18
18 = 6x
3 = x
Answers: 3;5;7

so you have
x
x+1
and
X+2

the product of the smaller two is 34 less the the square of the lager number

x(x+1)= -34 +(x+2)^2

kicks out x is equal to 6

so the answer is 6, 7, 8

The meaning of consecutive odd numbers implies that the common difference between any two adjacent numbers is 2. Let x-1, x+1 ,and x+3 be the three numbers.
(x-1)(x+1) = (x+3)^2 - 34
=>x^2-1 = x^2+6x-25
=>6x = 24
x = 4
So, the 3 consecutive odd numbers are: 3, 5, 7

Let n, n+2 and n+4 be the cosecutive ODD numbers.
So, n(n+2) + 34 = (n+4)^2.
solve this.
2n + 34 = 8n + 16
18 = 6n
n = 3.
So 3, 5,7 are the numbers.
Happy?

suppose the smallest odd number was x
the second number would be x + 2, and the third number would be x + 4
so the product of the two smaller numbers (x, x + 2) is 34 less than the square of the largest number (x + 4)
x(x+2) = (x+4)^2 - 34
x^2 + 2x = x^2 + 8x + 16 - 34
6x = 18
x = 3
so the three number are 3, 5, and 7

3, 5, 7

3x5=15
7^2=49
49-15=34