x^4-6x^3+7x^2-6x+6=x^4-6x^3+6x^2 + x^2-6x+6=x^2(x^2-6x+6)+(x^2-6x+6)=
(x^2+1)(x^2-6x+6)
x^2+1 has no real roots; to find roots of x^2-6x+6=x^2-6x+9-3=0, so (x-3)^2=3, x-3=+sqrt(3), and x=3+sqrt(3), x=4.7321 and 1.2679 approx.
Therefore, the real roots are 3+sqrt(3) and 3-sqrt(3) or 4.7321 and 1.2679.