All possible rational solutions are ± divisors of 4
These are

±1, ±2, ±4

We begin by trying 1. To use synthetic division we must 
consider the equation to be:



1 | 1  0 -6  4
  |    1  1 -5
    1  1 -5 -1

Nope. The remainder is -1 not 0, so 1 is not a solution.

We try -1:

-1 | 1  0 -6  4
   |   -1  1  5
     1 -1 -5  9

Nope. The remainder is 9 not 0, so -1 is not a solution.

We try 2:

 2 | 1  0 -6  4
   |    2  4 -4
     1  2 -2  0

Yep! The remainder is 0, so x=2 is a solution.  Therefore
we have factored the left side as



Using the zero-factor principle:

Setting  of course gives the solution we just found,
namely 

Setting 



This does not factor, so we use the quadratic formula
with , , 



















, , and 

That's 3 real solutions.  No imaginary solutions.

Edwin