Complete the square for $f(x) = 2x^2 + 4x - 6$ I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$
But I get stuck here. Can someone complete it and explain?
I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

Given: 

$f(x) = 2x^2 + 4x - 6$. complete the square.

$f(x)=2x^2+4x-6$
$f(x)=2x^2+4x-6 $ 
$f(x)=2(x^2+2x)-6$ 
$f(x)=2[x^2+\dfrac{2x}{2}+(1)^2]-6$
$f(x)=[2(x+1)^2 -1]-6$ 
$f(x)=2(x+1)^2-8$  which is the same as: $f(x) = 2x^2 + 4x - 6$
This can be confirmed by using a graphing calculator such as: https://www.desmos.com/
