Question 1029126

Let {{{x}}},{{{(x+2)}}},{{{(x+4)}}} represent the three consecutive odd integers. 

if {{{3}}} times the {{{sum}}} of all three is {{{28}}} more than the {{{product}}} of the first and second integers, we have:

{{{3(x+(x+2)+(x+4))=x(x+2)+28}}}.....solve for {{{x}}}

{{{3(x+x+2+x+4)=x^2+2x+28}}}

{{{3(3x+6)=x^2+2x+28}}}

{{{9x+18=x^2+2x+28}}}

{{{x^2+2x-9x-18+28=0}}}

{{{x^2-7x+10=0}}}

{{{x^2-5x-2x+10=0}}}

{{{(x^2-5x)-(2x-10)=0}}}

{{{x(x-5)-2(x-5)=0}}}

{{{(x-5) (x-2) = 0}}}


solutions:
{{{(x-5) = 0}}}->{{{x=5}}}
or
{{{(x-2) = 0}}}->{{{x=2}}}->even

so, your numbers are:{{{5}}},{{{7}}},{{{9}}}