Question 779012
{{{x^2 + (2x + 4)^2 = (2x + 6)^2}}}....using the Pythagorean theorem {{{a^2+b^2=c^2}}}, means we have right angle triangle where side {{{x=a}}}, side {{{b=(2x + 4)}}}, and hypotenuse {{{c=(2x + 6)}}}

so, first find {{{x}}}

{{{x^2 + (2x)^2 +16x+4^2 = (2x)^2 +24x+ 6^2}}}

{{{x^2 + 4x^2 +16x+16 =4x^2 +24x+ 36}}}...simplify

{{{x^2 +cross( 4x^2) +16x+16 =cross(4x^2) +12x+ 36}}}

{{{x^2 +16x+16 =24x+ 36}}}

{{{x^2 +16x+16-24x-36=0}}}

{{{x^2-8x-20=0}}}.....use quadratic formula to find {{{x}}}

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-(-8) +- sqrt( (-8)^2-4*1*(-20) ))/(2*1) }}}

{{{x = (8 +- sqrt( 64+80 ))/2 }}}

{{{x = (8+- sqrt( 144))/2 }}}

{{{x = (8+- 12)/2 }}}

solutions:

{{{x = (8 + 12)/2 }}}

{{{x = 20/2 }}}

{{{x = 10 }}}

or

{{{x = (8 - 12)/2 }}}

{{{x = -4/2 }}}

{{{x = -2 }}}

now, if we need to find the length of sides, we will use only positive solution
{{{x = 10 }}}


{{{x=a}}}=> {{{a=10 }}}

{{{b=(2x + 4)}}}=>{{{b=(2*10 + 4)}}}=> {{{b=24}}}

{{{c=(2x + 6)}}}=>{{{c=(2*10 + 6)}}}=>{{{c=26}}}

check:

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

{{{10^2+24^2=26^2}}}

{{{100+576=676}}}

{{{676=676}}}