Question 228601
Let x = second number
x+3 = first number


Sum of squares IS  37 more than the product of the numbers.

x^2 + (x+3)^2= x(x+3) + 37
x^2 + x^2 + 6x +9 = x^2 +3x + 37


Set equation equal to zero by subtracting x^2 +3x + 37 from each side:
2x^2 + 6x + 9-x^2 -3x-37=0
x^2 + 3x - 28=0


This just happens to factor!  
(x+7)(x-4)=0


There are two solutions:


First solution:
x=-7   
x+3= -4


Second solution:
x=4
x+3= 7


Both solutions do check if you take the time to do it.


Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus