Help..Completing the square problem

Solve 2x^2-12x+4=0 by competing the square, expressing the result in simplest radical form

2x^2 -12x +4 =0

2(x^2 -6x) = -4

2(x^2 -6x +3^2 -3^2) = -4

2(x^2 -6x +9) + 2(-9) = -4

2(x-3)(x-3) -18 = -4

2(x-3)^2 = -4 +18

2(x-3)^2 = 14

(x-3)^2 = 14/2

x - 3 = +/- √7

x= 3 +/- √7

can't simplify further.

2x^2-12x+4=0
2(x^2-6x+2)=0
2[(x-3)^2-7]=0
Finally we get
2(x-3)^2-14=0