Question 421777
using the pythagoras theorem

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

x^2+4x^2+8x+4=4x^2+12x+9
x^2-4x-5=0
x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x= 5 

The sides are
one leg  5 in

second (2x+2)===>12 in
hypotenuse (2x+3)===>13 in