Answers

2016-07-09T21:07:15+08:00
Let x - second number, and x + 3 - the first number.
(x)^2 + (x + 3)^2 = 37 + (x)(x + 3)
x^2 + x^2 + 6x + 9 = 36 +(x^2 + 3x)
2x^2 + 6x + 9 = x^2 + 3x + 37
x^2 + 3x - 28 = 0
(x + 7)(x - 4) = 0
x = -7 and x = 4

Testing the value, we have
-7 and -4 
49 + 16 = 65
(-7)(-4) + 37 = 28 + 37 = 65

4 and 7
16 + 49 = 65
(4)(7) = 28 + 37 = 65

Therefore, the numbers can be -7 and -4, or 4 and 7.

0