Question 3696
 f(x) = x^3 - 6x - 9
       = x^3 - 3x^2 + 3x^2 - 
           6x - 9

         = x^2(x-3) + 3(x^2 -2x -3)
         = x^2(x-3) + 3(x-3)(x+1)
         = (x-3)(x^2 + 3x +3)
 So, x -3=0 or x^2 +3x+3 = 0
 By quadratic formula
  x = (-3+ sqrt(9-12)  )/2 = (-3 + sqrt(3) i)/2 
 or (-3 - sqrt(3))/2 

( The approximated values are -1.5 -0.87 i or 
  -1.5 +0.87 i ,to the nearest hundreds.)
I don't think to the tenth is good enough.)

 Kenny