Question 87922
Let the first integer be x.
Second integer will be 2x + 1.
Third integer will be {{{x^2 - 3}}}

It is given that sum of three positive integers is 68.


so, {{{x + (2x+ 1) + (x^2 - 3) = 68}}}

{{{x^2 + 3x - 2 = 68}}} 

After simplification we will get
{{{x^2 + 3x - 70 = 0 }}}

Factorize it, we will get

(x+10)(x-7) = 0

x = -10, 7

As the integer is positive so will take the positive value

Therefore the first integer will be 7
Second integer = 2x + 1 = 2*7 + 1 = 15
Third integer = {{{x^2 - 3 = 7^2 - 3 = 46}}}