Question 846776
Use the Pythagorean theorem,
{{{(x+3)^2=(x-5)^2+(x+2)^2}}}
{{{x^2+6x+9=x^2-10x+25+x^2+4x+4}}}
{{{x^2+6x+9=2x^2-6x+29}}}
{{{x^2-12x+20=0}}}
{{{(x-2)(x-10)=0}}}
Two solutions:
{{{x-2=0}}}
{{{x=2}}}
The sides would then be -3,0,and 5.
This is not a good solution.
{{{x-10=0}}}
{{{x=10}}}
The sides would then be 5,12,and 13.
This is a good solution.