Question 740931
sides: x, 2x+4=2(x+2), 2x+6=2(x+3)

so we write:


{{{x^2+(2(x+2))^2=(2(x+3))^2}}}

{{{x^2+4(x^2+4x+4)=4(x^2+6x+9)}}}

{{{x^2+4(x^2+4x+4)-4(x^2+6x+9)=0}}}


{{{x^2-8x-20=0}}}

{{{x^2-10x+2x-20=0}}}

{{{x(x-10)+2(x-10)=0}}}

{{{(x-10)(x+2)=0}}}

which gives us x=-2  or x=10

but x>0


so only x=10 works   :)