Question 845886
Use the Pythagorean theorem.
{{{(x+2)^2+(4x)^2=(4x+1)^2}}}
{{{x^2+4x+4+16x^2=16x^2+8x+1}}}
{{{x^2-4x+3=0}}}
{{{(x-3)(x-1)=0}}}
Two solutions:
{{{x-3=0}}}
{{{x=3}}}
The sides are then 5,12,13.
{{{x-1=0}}}
{{{x=1}}}
The sides are then 3,4,5.