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  "body": "\n## Statement\n\nLet $a$, $b$ and $c$ be complex numbers with $a \\neq 0$, and let $s$ be a complex number with $s^2 = b^2 - 4 a c$.\nThen, for each complex number $x$,\n$$a x^2 + b x + c = 0 \\iff x = \\frac{-b + s}{2 a} \\text{ or } x = \\frac{-b - s}{2 a}.$$\n\n## Remarks\n\nSuch an $s$ exists, by the [[existence of nth roots|0YM6P6C206]] for $n = 2$.\nThe number $\\Delta = b^2 - 4 a c$ is the discriminant. For real $a, b, c$, there are two real roots if $\\Delta \u003e 0$, one double real root if $\\Delta = 0$, and two conjugate non-real roots if $\\Delta \u003c 0$.\nThe same formula holds in each field in which $2 \\neq 0$.\n\n## Explanations\n\n### L1 human\n\nThe formula solves every quadratic equation.\nFor $x^2 + 2 x + 5 = 0$, the discriminant is $4 - 20 = -16$, so $s = 4 i$, and the roots are $-1 + 2 i$ and $-1 - 2 i$.\nCompleting the square proves the formula: the equation becomes $(2 a x + b)^2 = b^2 - 4 a c$.\n",
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