The answer can be found very quickly.
a2-4a-1=0; Let us divide both sides by a. Since we are to find a-1/a, a≠0, otherwise the expression a-1/a is not defined at all. After the abovementioned division by a we will obtain:
a-4-1/a=0; Now combine first and last term together, (a-1/a)-4=0. Move 4 to the right, to obtain
This is your answer. Done.
In order to solve for "a - (1/a)" (or "(a-1)/a" as what you have can be read either way), you must first know the value of a. To do this, the equation a^2 - 4a - 1 = 0 must be solved for a.
To solve for a in this quadratic equation, we can do one of two things.
So let's solve by completing the square. I give both your example and a generalized formula using your example's set up. First, we move the numerical value over to the right side of the equation. Next, we look at the b coefficient in the equation - half that number (its positive/negative sign included), square it, and add that new value to both sides of the equation. The left side of the equation can now be written as a perfect square - that's why we completed the square in the first place. Take the square root of both sides of the equation and use PEMDAS rules to solve for your variable. Keep in mind the +/- that goes with taking the square root of any number!
a^2 - 4a - 1 = 0 ax^2 - bx - c = 0 (assume "a" = 1 and "b" and "c" are neg)
a^2 - 4a - 1 + 1 = 0 + 1 x^2 - bx - c + c = 0 + c
a^2 - 4a = 1 x^2 - bx = c
a^2 - 4a + 4 = 1 + 4 x^2 - bx + (b^2)/4 = c + (b^2)/4
(a - 2)^2 = 5 (x - b/2)^2 = c + (b^2)/4
a - 2 = +/- sqrt(5) x - b/2 = +/- sqrt[c + (b^2)/4]
a - 2 + 2 = +/- sqrt(5) + 2 x - b/2 + b/2 = +/- sqrt[c + (b^2)/4] + b/2
a = 2 +/- sqrt(5) x = b/2 +/- sqrt[c + (b^2)/4]
Double checking the two solutions: a = 2+sqrt(5) and a = 2-sqrt(5), both work in the original equation. Since the equation has two answers, your desired value a-1/a will have two answers as well. Just substitute both values (one at a time - never together) or plug the values into a calculator to get your final results.
Note: my general example contains a few assumptions. You can play around using the basic quadratic formula to see how different combinations will work out. In fact, doing so using just the standard ax^2+bx+c=0 will lead you to the quadratic formula (solving option #2).