Question 262218
Let x, x+2, x+4 be our 3 consecutive odd integers.
from above, we get
{{{x*(x+2) = x+4 + 8}}}
distributing the left we get
{{{x^2 + 2x = x + 4 + 8}}}
simplifying the right we get
{{{x^2 + 2x = x + 12}}}
setting the equation = 0, we get
{{{x^2 + x - 12 = 0}}}
factoring we get
{{{(x+4)(x-3) = 0 }}}
solving each for x we get
{{{x = -4}}} - which we can't use
and
{{{x = 3}}}
this means we have 
{3, 5, 7}