Quadratic Formula

Since \(z=\pm\frac{\sqrt{b^24ac}}{2a}\), would that make \(r=\frac{b}{2a}\pm\frac{\sqrt{b^24ac}}{2a}\)？

@divinedolphin Yes! That's essentially the same thing as the quadratic formula. The only tiny difference is the "\(\textcolor{red}{ \pm } \)" which simplifies to the normal "\(\textcolor{red}{\pm}.\)
\(r=\frac{b}{2a}\textcolor{red}{\pm}\frac{\sqrt{b^24ac}}{2a}\)
same as\(r=\frac{b}{2a}\textcolor{red}{\pm}\frac{\sqrt{b^24ac}}{2a}\)

(Latex didn't work so here's a picture:)

@tidyboar
By the way, here is the code for the quadratic formula above\\(r=\frac{b}{2a}\textcolor{red}{\pm}\frac{\sqrt{b^24ac}}{2a}\\)
\(r=\frac{b}{2a}\textcolor{red}{\pm}\frac{\sqrt{b^24ac}}{2a}\)