Question 1179118
If x is one number the other is x+5

The sum of their squares is 157
x^2+(x+5)^2=157

x^2+x^2+10x+25=157

2x^2+10x-132=0
divide by 2

x^2+5x-66=0

x^2+11x-6x-66 =0

x(x+11)-6(x+11)=0

(x+11)(x-6)=0

x=-11, or x=6

when one number is -11 the is -6   
when the number is 6 the othr is 11