SOLUTION: Find all solutions, real or complex, to the following equation {{{ x^3 - 4x^2 - 9x + 36=0 }}}

  x³-4x²-9x+36

You called that an equation, but there is no equal sign, so
it isn't an equation.  Was it supposed to have = 0 after it
like this?:

  x³-4x²-9x+36 = 0

Then it would have been an equation to solve.  But if it
is only

  x³-4x²-9x+36

then we can only factor it, not solve it.  

First I will assume there was no = 0 after it and the 
instructions were not to solve the equation but to 
factor the expression

Factor the first two terms x³-4x² by taking out the
greatest common factor, x², getting x²(x-4)

Factor the last two terms -9x+36 by taking out the
greatest common factor, getting -9(x-4)

So we have

x²(x-4)-9(x-4)

Notice that there is a common factor, (x-4)

x²(x-4)-9(x-4)

which we can factor out leaving the x² and the -9 to put 
in parentheses:

(x-4)(x²-9)

Notice that the (x²-9) is the difference of two perfect squares

So the final factorization is

(x-4)(x-3)(x+3)

That's the final answer if the instructions were "factor the 
expression".

However if it was an equation as you stated and there was an equal 
sign and a 0 after it, like this:

(x-4)(x-3)(x+3) = 0

then we use the zero factor principle:

x-4=0   x-3=0   x+3=0
  x=4     x=3     x=-3

Then there are three solutions to the equation, and

they are 4,3,and -3.

Edwin