Use synthetic division:
I will write it up a little verbosely to make the process as clear as possible. Essentially it's like doing long division.
First notice that 3x(2x - 1) = 6x^2-3x, which will allow us to separate the x^2 term:
(6x^2 - x + 6) - 3x(2x - 1)
= (6x^2 - x + 6) - (6x^2-3x)
= 6x^2 - x + 6 - 6x^2 + 3x
=2x + 6
So 3x makes a good multiplier, with remainder #(2x+6).
To match the leading term 2x in the remainder, the next multiplier we want is 1:
1xx(2x-1) = 2x-1
Then
(2x + 6) - (2x - 1) = 2x + 6 - 2x + 1 = 7
This gives 7 as our final remainder. If the remainder was zero then we would have a simple factorization.
Add the multipliers that we have found together to get the term (3x+1).
So to summarize where we have got to so far:
(6x^2 - x + 6) = (3x + 1)(2x - 1) + 7
Therefore
(6x^2 - x + 6)/(2x - 1) = (3x + 1) + 7/(2x - 1)
= 3x + 1 + 7/(2x - 1)