x³ - 7x² + 19x - 13 

The only possible zeros are the factors of 13, which are ±1 and ±13.

So we use synthetic division to try the easiest one first, 1:

1 | 1 -7  19 -13
  |    1  -6  13
   -------------
    1 -6  13   0

We're in luck because that left a 0 remainder.
So we have factored the polynomial as

(x - 1)(x² - 6x + 13)

It has zeros found by setting each of those factors = 0

The first one gives 1 as a zero.  The other two zeros
are found by the quadratic formula.

They are 3±2i

Edwin