Question 1152201

Let the numbers are {{{x}}},{{{x+1}}},{{{x+2}}}

if the product of the first and the second is {{{37}}} less than the square of the third, we have

{{{x(x+1)=(x+2)^2-37}}}

{{{x^2+x=x^2+4x+4-37}}}

{{{cross(x^2)+x=cross(x^2)+4x-33}}}

{{{x=4x-33}}}

{{{33=4x-x}}}

{{{33=3x}}}

{{{x=11}}}


the numbers are: {{{11}}},{{{12}}},{{{13}}}