[billyjmc]

What you’ve suggested is not equivalent. The a is always taken without negation, and that’s not the case in your rendering.

[LordDragonfang]

Ah. "a±(b-c)" then.

The fact I made that mistake in the first place just makes me more convinced the original form is unnecessarily complex to parse.

[stabbles]

You're too fixated on the example. It's obviously used in cases where you cannot group like that:

    (-b ± sqrt(b² - 4ac)) / 2a = 2c / (-b ∓ sqrt(b² - 4ac))