Question 154288
Let x be the length of the shorter leg, then that of the longer one is x + 3.
Applying the Pythagorean Theorem, we have
{{{x^2 + (x+3)^2=23^2}}}
Solving for x, we have
{{{x^2+x^2+6x+9=529}}}
{{{2x^2+6x-520=0}}}
{{{x^2+3x-260=0}}}   (divide both sides by 2)
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}    (plug in a=1,b=3,c=-260)
{{{x = (-(3) +- sqrt( 3^2-4*1*(-260) ))/(2*1) }}}
{{{x = (-3 +- sqrt( 9+1040 ))/2 }}}
{{{x = (-3 +- sqrt( 1049 ))/2 }}}
So 
{{{x = (-3 + sqrt( 1049 ))/2 = 14.69}}}
Or
{{{x = (-3 - sqrt( 1049 ))/2=-17.69 }}}  (reject this solution as the length can not be negative)
Thus the length of the shorter leg is 14.69 units, and the length of the longer one is x+3=14.69+3=17.69 units