Question 1003673

let the three consecutive positive integers be {{{x}}},{{{x+1}}},and {{{x+2}}}

if the product of two consecutive positive integers is {{{119}}} more than the next integer, we have 

{{{x(x+1)=(x+2)+119}}}

{{{x^2+x=x+121}}}

{{{x^2+x-x=121}}}

{{{x^2=121}}}

{{{x=sqrt(121)}}}........since given that integers are positive

solution we need is:{{{x=11}}}

so, 
first integer is:{{{11}}}
second integer is:{{{12}}}
third integer is:{{{13}}}=>the largest of the three integers