Question 748392
let two factors of {{{-36}}} be {{{x}}} and {{{y}}} 

if one factor is {{{11}}} less than half of the other factor, we have

{{{x+11=y/2}}} => {{{y=2x+22}}}
and {{{xy=-36}}}

{{{x(2x+22)=-36}}}

{{{2x^2+22x+36=0}}}

{{{x^2+11x+18=0}}}.....write {{{11x}}} as {{{2x+9x}}}
{{{x^2+2x+9x+18=0}}}
{{{(x^2+2x)+(9x+18)=0}}}
{{{x(x+2)+9(x+2)=0}}}
{{{(x+9)(x+2)=0}}}

solutions:

{{{x=-9}}}
{{{x=-2}}}

then

{{{y=2x+22}}} => {{{y=2(-9)+22}}} => {{{y=-18+22}}} => {{{y=4}}}
or
{{{y=2x+22}}} => {{{y=2(-2)+22}}} => {{{y=-4+22}}} => {{{y=18}}}

so, pairs are: 

{{{x=-9}}} and {{{y=4}}}

or {{{x=-2}}} and {{{y=18}}}


the choices are
{{{-2}}} and {{{18}}},


we can check these choices too:

{{{-6}}} and {{{6}}},....plug in {{{y=2x+22}}}

{{{6=2(-6)+22}}}

{{{6=-12+22}}}
{{{6<>10}}}......not a solution


{{{3}}} and {{{-12}}},....plug in {{{y=2x+22}}}

 {{{-12=2*3+22}}}


 {{{-12<>28}}}.....not a solution

{{{4}}} and {{{-9}}}....plug in {{{y=2x+22}}}

{{{-9=2*4+22}}}

{{{-9<>30}}}.....not a solution


so, your solution is {{{-2}}} and {{{18}}}