Let the first number = x
Let the second number = y

So FIRST NUMBER MINUS SECOND NUMBER EQUALS 8

That is,

x - y = 8

(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