Take 2 as a common factor out of the first 2 terms
2(x^2 + 6x) + 3 = 0
Add half 6 squared inside the brackets. Subtract the same number outside the brackets when multiplied by 2
2(x^2 + 6x + 3^2) + 3 - 2*3^2 = 0
Combine the two terms on the end.
2(x^2 + 6x + 9) + 3 - 2*9 = 0
2(x^2 + 6x + 9) + 3 - 18 = 0
2(x^2 + 6x + 9) - 15 = 0
Express the terms inside the brackets as a perfect square
2(x + 3)^2 - 15 = 0
Discussion Graph
Technically, the vertex form would y = 2(x + 3)^2 - 15. It needs to be more general than equated to zero. I have solved it the way you presented it and graphed it by definition.
The red graph is y = 2x^2 + 12x + 3
The blue graph is y = 2(x + 3)^2 - 15
Notice they have the same critical point which means they are equivalent