How to solve the system of equations algebraically over a set of real numbers

Solve the following equation in the real number system. Please show all of your work.
example: 
x³-6x²-10x+12=0
...
f(x)=x³-6x²-10x+12
A graphing calculator will show that there is one zero at x=-2
Divide f(x) by (x+2) by long division or synthetic division
(x³-6x²-10x+12)/(x+2)=x^2-8x+6

solve for other two zeros
x^2-8x+6=0
complete the square
(x^2-8x+16)+6-16=0
(x-4)^2=10
x-4=±√10
x=4±√10=7.16 and -.84
ans:
x=-2, -.84 and 7.16

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.