How do you find the domain of #f(x)=(x-3)/(x^2-x-2)#?

To find the domain of a rational function, you need to determine all values of #x# for which the demoninator is equal to #0#.

So, set #x^2 - x - 2 = 0# and solve the quadratic equation.

There are several possibilities to solve a quadratic equation, one of my favourite is "completing the circle":

#x^2 - x - 2 = 0#
#<=> x^2 - x = 2#

Try to write the left side of the equation like #(x-a)^2 = x^2 - 2ax + a^2#.
We already have the #x^2#, and from #-x = -2ax# we can conclude that #a = 1/2#. So, the only term that is missing on the left side to complete the quadratic form is #+1/4#.

In order to prevail the equality, the term needs to be added on both sides of the equation, leading us to the following:
# x^2 - x + 1/4 = 2 + 1/4#
#<=> (x-1/2)^2 = 9/4#

Now, we can calculate the root on both sides. Don't forget that the root of #9/4# has two solutions since both #(3/2)^2 = 9/4# and #(-3/2)^2 = 94# hold.

Thus, we can further solve the quadratic equation as follows:
#x - 1/2 = 3/2 or x - 1/2 = - 3/2#
#<=> x = 2 or x = -1#

This means that for those two values, the denominator of our function would be #0# which is not admissible.

So, the domain of the function are all real numbers except #x = 2# and #x = -1#.
# x in RR# \ #{-1; 2}#