Difference between revisions of "User talk:Meljel"

(It's me meljel lol)
(It's me meljel)
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== It's me meljel lol ==
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== hi. ==
  
<cmath> \begin{align*}
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if ur from hysb then bye
ax^2 + bx + c &= 0\\
 
a(x^2 + \frac bax + \frac ca) &= 0\\
 
a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\
 
{(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\
 
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\
 
{(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\
 
{(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\
 
x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\
 
x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\
 
x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\
 
x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\
 
\end{align*}
 
</cmath>
 
  
 
== hihiihi ==
 
== hihiihi ==

Revision as of 18:04, 16 January 2021

hi.

if ur from hysb then bye

hihiihi

\[a^2-b^2=(a+b)(a-b)\] \[(\cos \theta + i\sin \theta)^n = \cos n\theta + ii\sin n\theta\]

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