Question 556089
Three consecutive integers. If we say the first one is n, than the 3 will be n, n+1 and n+2
Therefore we can write the problem as
n*(n+1)=-6*(n+2)
Distribute the n and -6
{{{n^2+n=-6n-12}}}
Add 6n to both sides
{{{n^2+7n=-12}}}
add 12 to both sides
{{{n^2+7n+12=0}}}
factor
(n+3)(n+4)=0
n=-3 or -4

So our integers would be -3,-2,-1
or -4,-3,-2

-3*-2=-1*6 
6=6

-4*-3=-2*-6
12=12