Difference between revisions of "User talk:Meljel"

Revision as of 00:49, 8 January 2019

It's me meljel lol

$\begin{align*} ax^2 + bx + c &= 0\\ a(x^2 + \frac bax + \frac ca) &= 0\\ a({(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca) &= 0\\ {(x + \frac {b}{2a})}^2 - \frac{b^2}{4a^2} + \frac ca &= 0\\ {(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac ca\\ {(x + \frac {b}{2a})}^2 &= \frac{b^2}{4a^2} - \frac {4ac}{4a^2}\\ {(x + \frac {b}{2a})}^2 &= \frac{b^2 - 4ac}{4a^2}\\ x + \frac {b}{2a} &= \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}\\ x + \frac {b}{2a} &= \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\ x &= - \frac {b}{2a} \pm\frac{\sqrt{b^2 - 4ac}}{2a}\\ x &= \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\\ \end{align*}$

hihiihi

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

@@ Line 16: / Line 16: @@
 </cmath>
-== It's me meljel lol ==
+== hihiihi ==
-hi
+<cmath>a^2-b^2=(a+b)(a-b)</cmath>
+<cmath>(\cos \theta + i\sin \theta)^n = \cos n\theta + ii\sin n\theta</cmath>

Page

Toolbox

Search

Difference between revisions of "User talk:Meljel"

Revision as of 00:49, 8 January 2019

It's me meljel lol

hihiihi