<PRE><B>Question 1171166</B>

let consecutive integers be {{{x}}} and {{{x+1}}}

A positive integer squared: {{{x^2}}}
plus 
6 times its consecutive integer: {{{6(x+1)}}}

is equal to {{{22}}}

put all together:

{{{x^2+6(x+1)=22}}}....solve for {{{x}}}

{{{x^2+6x+6-22=0}}}

{{{x^2+6x-16=0}}}....factor

{{{(x - 2) (x + 8) = 0}}}


solutions:

{{{x=2}}}
or
{{{x=-8}}}


consecutive integers are {{{2}}} and {{{3}}}
or
consecutive integers be {{{-8}}} and {{{-7}}}

</PRE>