Step 1: move pieces around and divide by the leading coefficient to put equation into the form x² + b = 2ax (or if you like, x² – 2ax + b = 0; or feel free to swap the sign of b if you prefer). The equation for the parabola is then x² + b = 2ax + y.

Step 2 (optional): rearrange that equation to get (x – a)² = a² – b

Step 3: x = a ± √(a² – b)

In this form, the “discriminant” is just a² – b, the x coordinate of the vertex is a and the y coordinate is b – a², Viète’s formulas tell us that the two roots satisfy ½(x₁ + x₂) = a and x₁x₂ = b. If the coefficients are real but the roots are complex, then we know each root has amplitude √b and phase arccos(a/√b). Etc.