Question 823899
let legs be {{{a}}} and {{{b}}}, and the hypotenuse {{{c}}}

let shorter leg be {{{a=x}}}

then the longer leg  is {{{b=a+3}}} or {{{b=x+3}}} 

if the hypotenuse {{{c}}} is {{{3}}} millimeters longer than the longer leg {{{b}}}, then {{{c=b+3}}}.....substitute {{{b=x+3}}} and you have {{{c=x+3+3=x+6}}}

use Pythagorean theorem:

{{{c^2=a^2+b^2}}}

{{{(x+6)^2=x^2+(x+3)^2}}}................solve for {{{x}}}

{{{x^2+12x+36=x^2+x^2+6x+9}}}

{{{x^2+12x+36=2x^2+6x+9}}}

{{{0=2x^2+6x+9-x^2-12x-36}}}

{{{x^2-6x-27=0}}}............write {{{-6x}}} as {{{-9x+3x}}} 

{{{x^2-9x+3x-27=0}}}............group

{{{(x^2-9x)+(3x-27)=0}}}............factor out {{{x}}} and {{{3}}}

{{{x(x-9)+3(x-9)=0}}}


{{{(x+3)(x-9)=0}}}

solutions:

if {{{(x+3)=0}}} => {{{x=-3}}}...........leg cannot be negative length, so disregard this solution

if {{{(x-9)=0}}}=> {{{x=9}}}..............your solution; now use it to find out the length of legs and hypotenuse

{{{a=x}}}=>{{{a=9}}}

{{{b=x+3}}} =>{{{b=9+3}}} =>{{{b=12}}} 

{{{c=x+6}}}=>{{{c=9+6}}}=>{{{c=15}}}


check our solution:

{{{c^2=a^2+b^2}}}

{{{15^2=9^2+12^2}}}

{{{225=81+144}}}

{{{225=225}}}