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————杨绛


When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
如果\(a \ne 0\),则方程\(ax^2 + bx + c = 0\)有两个根,他们是 $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

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