Since
is the square of x and 6x is twice the product between x and 3, the second square must be 3 squared, i.e. 9.
So, if we think of 15 as 9+6, we have

Which is the required vertex form. This form tells us imediately that the vertex is the point (3,6).
Since the leading coefficient is 1, the parabola is facing upwards (it's U shaped), so the vertex is a minimum.