Question 468907
Let x - the shortest leg of a triangle
The shortest leg of a triangle is 7 inches shorter than the other leg, then the other leg is {{{x+7}}} inches. The hypotenuse of this triangle is 13 inches.
Use the Pythagorean theorem {{{x^2+(x+7)^2=13^2}}}
{{{x^2+x^2+14x+49=169}}}
{{{2x^2+14x-120=0}}} Divide by 2
{{{x^2+7x-60=0}}} 
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-7 +- sqrt( 7^2-4*1*(-60) ))/(2*1) }}} 
{{{x = (-7 +- sqrt( 289))/2 }}} 
{{{x = (-7 +17)/2 =5}}} inches
{{{x = (-7 -17)/2 =-12<0}}} extraneous root
the shortest leg of a triangle is 5 inches, the other leg is 5+7=12 inches