Question 813188

let {{{3}}} consecutive positive integers be:
 {{{x}}}-first,
{{{(x+1)}}}-second,and 
{{{(x+2)}}}-third

if the product of the first and the third is {{{29}}} more than the second, then we have

 {{{x(x+2)=(x+1)+29}}}.....solve for {{{x}}}

{{{x^2+2x=x+1+29}}}

{{{x^2+2x=x+30}}}

{{{x^2+2x-x-30=0}}}

{{{x^2+x-30=0}}}.........write {{{x}}} as {{{6x-5x}}}

{{{x^2+6x-5x-30=0}}}.....group

{{{(x^2+6x)-(5x+30)=0}}}

{{{x(x+6)-5(x+6)=0}}}

{{{(x-5)(x+6) = 0}}}

solutions:

if {{{(x-5)  = 0}}}=>{{{x=5}}}

{{{ (x+6) = 0}}}=>{{{x=-6}}}.....since we need consecutive positive integers, disregard this solution

so, your consecutive positive integers are:
{{{x=5}}}
{{{x+1=6}}}
{{{x+2=7}}}