Question 553647
Let {{{x}}} be the length (in inches) of the shorter leg.
Then, the length (in inches) of the other leg would be
{{{x+6}}}
Pythagoras' theorem, applied to that triangle, tells you that
{{{(x+6)^2+x^2=25^2}}} --> {{{x^2+12x+36+x^2=25^2}}}
--> {{{2x^2+12x+36-25^2=0}}} --> {{{2x^2+12x-589=0}}}
Using the quadratic formula we find 2 values for {{{x}}}
{{{x = (-12 +- sqrt(12^2-4*2*(-589) ))/(2*2)=(-12 +- sqrt(4856))/4}}} ,
but we can only use the positive solution
{{{x =(-12 +- sqrt(4856))/4=-3 +- sqrt (1214)/2}}} = approximately {{{14.42125}}} .
Then, the other leg length is about {{{6+14.42125=20.42125}}} .
So the leg lengths, to the nearest inch, are 14 inches and 20 inches.