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| − | == It's me meljel lol == | + | == hi. == |
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| − | <cmath> \begin{align*}
| + | if ur from hysb then bye |
| − | ax^2 + bx + c &= 0\\
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| − | a(x^2 + \frac bax + \frac ca) &= 0\\
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| − | a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\
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| − | {(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\
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| − | {(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\
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| − | {(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\
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| − | {(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\
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| − | x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\
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| − | x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\
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| − | x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\
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| − | x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\
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| − | \end{align*}
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| − | </cmath>
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| | == hihiihi == | | == hihiihi == |
![\[a^2-b^2=(a+b)(a-b)\]](http://latex.artofproblemsolving.com/2/4/b/24b8beb8dd50e02f82066210105bb11abe192e0f.png)
![\[(\cos \theta + i\sin \theta)^n = \cos n\theta + ii\sin n\theta\]](http://latex.artofproblemsolving.com/4/6/c/46c3219c82358bf2b221c3738e42f52bdc422f5b.png)
