Question 857368
 
 
For three consecutive integers, 
Let X = 1st
X+1 = 2nd
X+2 = 3rd
Therefore, the sum of the three is X+(X+1)+(X+2), or 3X+3 total.
For the equation, twice the sum is 3 times the product of the larger two, or
{{{2(3X+3)=3(X+1)(X+2)}}}
From here, distribute
{{{6X+6=(3X+3)(X+2)}}}
Then, distribute some more...
{{{6X+6=3X^2+9X+6}}}
Subtract (6X+6)
{{{0=3X^2+3X+0}}}
Then, use the quadratic formula to find your two X's
{{{X=(-3+-sqrt(3^2-4*3*0))/(2*3)}}}
Do the appropriate multiplication
{{{X=(-3+-sqrt(9-0))/(6)}}}
Keep simplifying
{{{X=(-3+3)/6}}}or{{{X=(-3-3)/6}}}
So the answer is either{{{X=(-3+3)/6=0}}} or {{{X=(-3-3)/6=-6/6=-1.}}}
 
And so, the two answers for X are 0 and -1. The second is X + 1, third is X + 2.
 
Your 2 answers are 0, 1, 2 or -1, 0, 1.
 
To check, plug each of them in:
{{{2(0+1+2)=3(1*2)}}}
{{{2(3)=3(2)}}}
{{{6=6}}}
Next, try -1, 0, 1:
{{{2(-1+0+1)=3(0*1)}}}
{{{2(0)=3(0)}}}
{{{0=0}}}
 
 
Your final, checked answers are 0, 1, and 2; or -1, 0, and 1.