Completing the square in a different form

Question

Say we had the function $y = -6x^2-3$ and wanted to find the turning point, my first goal is to complete the square, but this is where it all falls apart. Of course I know you can differentiate both sides with respect to $x$ and get $y'=-12x$, which we can solve by setting it equal to $0$ and get $x=0$ for our turning point, which is correct. However I would like to ask is there a way to get this into the normal "vertex" form that we have for quadratics in the form $ax^2+bx+c$ where $b$ isnt equal to $0$?

Answer 1

$y = -6x^2-3 = -6(x - [0])^2 + {-3}$.

The vertex is at $(0, -3)$.