The root of a function, is where the function would evaluate to 0. In other words, you find all the x values, for which y or better say f(x) would be zero. Since a fraction would evaluate to 0 only if the numerator is 0 (since 0 divided by anything would be 0), for finding the root we simply write:
2x-3=0 => 2x=3 => x=3/2
So x=3/2 is the x value for which your function would evaluate to 0, and thus it is the root of the function.
In this case, at x=3/2, your function would be:
f(3/2) = [2*(3/2) -3] / (2*(3/2)) = [3-3] / [3] = 0/3 = 0
Now, to see if there is a hole or vertical asymptote, you have to remember that a function is not defined at x values where the denominator evaluates to 0. That is because a number divided by 0 is undefined and thus it can possibly have a vertical asymptote. The only way it would not have a vertical asymptote is that you can simplify the numerator and the denominator. In this case, you can not simplify (2x-3) and 2x, so there would be a vertical asymptote at x=0 (because 2x=0 => x=0 which is the x value that makes denominator 0).
Now let's look at a case where you would have a hole instead of vertical asympotote.
let f(x) = (5* (X^2) - 6x) / (2x)
in this case, we can factor an x from the numerator, and write:
f(x) = [x(5x-6)] / (2x)
now we can cross out the x in numerator and denominator, and we are left with:
f(x) = (5x-6)/2 = 2.5x -3 x not equal to 0
Notice that we have a new expression for f(x) which is valid for all values of x, except for x=0 and that is because in the original f(x) which was a fraction, x=0 would make the denominator 0 and thus the function would be undefined at that point. So now we would not have a vertical asymptote, but rather a hole on the linear function f(x) =2.5x-3.
If you have a graphing calculator, it would be even more clear.
Kaveh
