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