Quadratic Formula
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Since \(z=\pm\frac{\sqrt{b^2-4ac}}{2a}\), would that make \(r=-\frac{b}{2a}-\pm\frac{\sqrt{b^2-4ac}}{2a}\)？
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@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 "\(\textcolor{red}{\pm}.\)
\(r=-\frac{b}{2a}\textcolor{red}{-\pm}\frac{\sqrt{b^2-4ac}}{2a}\)
same as\(r=-\frac{b}{2a}\textcolor{red}{\pm}\frac{\sqrt{b^2-4ac}}{2a}\)
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(Latex didn't work so here's a picture:)
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@tidyboar
By the way, here is the code for the quadratic formula above\\(r=-\frac{b}{2a}\textcolor{red}{-\pm}\frac{\sqrt{b^2-4ac}}{2a}\\)\(r=-\frac{b}{2a}\textcolor{red}{-\pm}\frac{\sqrt{b^2-4ac}}{2a}\)
