How do I solve 2x^2 -4x -6 =0 by complete the square method

Question

So I got the vertex witch is (1,-8) and I put it into vertex form y=2(x-1)^2-8 but I can’t figure out how to do the rest

Answer 1

1. Divide everything by 2. You now have x²-2x-3=0
2. Isolate the variable. x² - 2x = 3
3. Take half of the linear term, square it and add to both sides half of -2 is -1, square it, 1 and add to both sides x² - 2x + (-1)² = 3 + 1
4. Write as (x + 1/2b)² (x - 1)^{2 = 4}
5. Take square root of both sides x - 1 = 2 or x -1 = -2
6. Solve for x x = 3 or x = -1

Answer 2

From here add 8 to both sides and then divide by 4 which will isolate the (x-1)^2 term

We then have (x-1)^2 = 4

Now take the square root of both sides to give you x-1= sq root of 4

so x-1 can = 2. Gives x=3

or x-1 can = -2 Gives x=-1