Question 230918
Let x= first number
x+5= second number


Your equation will be 


{{{x^2+(x+5)^2=37}}}
{{{x^2 +x^2+10x+25=37}}}


{{{2x^2+10x-12=0}}}
{{{2(x^2+5x-6)=0}}}
{{{2(x+6)(x-1)=0}}}
x=-6 or x=1


There are two solutions:

x=-6 and x+5=-1


Also,
x=1 and x+5=6


Dr. Robert J. Rapalje