Question 333647
We basically have this triangle set up:



{{{drawing(500,500,-0.5,2,-0.5,3.2,
line(0,0,0,3),
line(0,3,2,0),
line(2,0,0,0),
locate(-0.2,1.5,x),
locate(1,-0.2,x+2),
locate(1,2,2x-2)
)}}}




Remember, the Pythagorean Theorem is {{{a^2+b^2=c^2}}} where "a" and "b" are the legs of a triangle and "c" is the hypotenuse.



Since the legs are {{{x}}} and {{{x+2}}} this means that {{{a=x}}} and {{{b=x+2}}}


   

Also, since the hypotenuse is {{{2x-2}}}, this means that {{{c=2x-2}}}.




{{{a^2+b^2=c^2}}} Start with the Pythagorean theorem.



{{{x^2+(x+2)^2=(2x-2)^2}}} Plug in {{{a=x}}}, {{{b=x+2}}}, and {{{c=2x-2}}} 



{{{x^2+x^2+4x+4=4x^2-8x+4}}} FOIL



{{{x^2+x^2+4x+4-4x^2+8x-4=0}}} Get every term to the left side.



{{{-2x^2+12x+0=0}}} Combine like terms.



Notice that the quadratic {{{-2x^2+12x+0}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=-2}}}, {{{B=12}}}, and {{{C=0}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(12) +- sqrt( (12)^2-4(-2)(0) ))/(2(-2))}}} Plug in  {{{A=-2}}}, {{{B=12}}}, and {{{C=0}}}



{{{x = (-12 +- sqrt( 144-4(-2)(0) ))/(2(-2))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144-0 ))/(2(-2))}}} Multiply {{{4(-2)(0)}}} to get {{{0}}}



{{{x = (-12 +- sqrt( 144 ))/(2(-2))}}} Subtract {{{0}}} from {{{144}}} to get {{{144}}}



{{{x = (-12 +- sqrt( 144 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{x = (-12 +- 12)/(-4)}}} Take the square root of {{{144}}} to get {{{12}}}. 



{{{x = (-12 + 12)/(-4)}}} or {{{x = (-12 - 12)/(-4)}}} Break up the expression. 



{{{x = (0)/(-4)}}} or {{{x =  (-24)/(-4)}}} Combine like terms. 



{{{x = 0}}} or {{{x = 6}}} Simplify. 



So the possible solutions are {{{x = 0}}} or {{{x = 6}}} 

  

Since a length of 0 doesn't make much sense, the only solution is {{{x = 6}}} 



So the legs of the triangle are 6 m and 8 m (add 2) while the hypotenuse is 10 m (double 6 and subtract 2)