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$2=\frac{b^2-4ac}{4a^2}$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

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

$x^2+px+q=0$

$\sin^2\theta + \frac{b}{\sqrt {ac}} \sin\theta \cos\theta \pm \cos^2\theta = 0$

By Editor CoAxial on 06 Aug 22.

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