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Question Number 215957 by LG last updated on 22/Jan/25

$$b^2-4ac$$

Commented by MathematicalUser2357 last updated on 23/Jan/25

$$x=\frac{-b\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2a}$$

Commented by ajfour last updated on 23/Jan/25

$$a{\left(\frac{-b\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2a}\right)}^2+b\left(\frac{-b\pm \sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2a}\right)+c=0$$

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