Question

Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8

Answer 1

If $x$ is the first integer, then the next two are $x+2$ and $x+4$.

If the product of the first and second are greater than the third integer by 8, that means you have the following equation:

$x(x+2)=(x+4)+8\implies x^2+x-12=(x-3)(x+4)=0\implies x=3\text{ or }x=-4$

Ignore $x=-4$ since it's an even integer.

Check that $x=3$ is correct. The next two odd integers would be $5$ and $7$. You have

$3\times5=15=7+8$

so this is correct.