Two numbers...  
Let the first number = x
Let the second number = y
differ by 8,...
So FIRST NUMBER MINUS SECOND NUMBER EQUALS 8
That is,
x - y = 8
and their product is 65...
(FIRST NUMBER) TIMES (SECOND NUMBER) EQUALS 65.
xy = 65
So we have the system of equations:
Solve the first equation for x
x-y = 8
  x = 8+y
Substitute (8+y) for x in the second equation:
    xy = 65
(8+y)y = 65
If you like, swap the two factors y and (8+y) on
the left.
y(8+y) = 65
Distribute to remove parentheses:
8y+y² = 65
Swap the terms on the left
y²+8y = 65
Subtract 65 from both sides:
y²+8y-65 = 0
Factor the left sides:
(y+13)(y-5) = 0
Use the zero-factor principle:
Set each factor = 0
y+13 = 0;    y-5 = 0
   y = -13     y = 5
For y = -13 we substitute in the equation
solved for y:
  x = 8+y
  x = 8+(-13)
  x = -5
So one solution is 
First number = x = -5
Second number = y = -13
There is another solution:
For y = 5 we substitute in the equation
solved for y:
  x = 8+y
  x = 8+(5)
  x = 13
So the other solution is 
First number = x = 13
Second number = y = 5
Edwin