Along the lines of thats its the only thing worth
>>8621105
>>8621232
u right dawg. Caint see shit.
[math] \displaystyle
f(x) = e^{-ix}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - e^{-i x}(i \cos x + i^2 \sin x) \equiv 0
\\
f^{\prime}(x) = 0 \;\;\; \forall \; x \in \mathbb{R}\Rightarrow f(x) \text{ is a constant}
\\
f(0) = e^{0}(\cos 0 + i \sin 0) = 1 \cdot(1+0) = 1 \Rightarrow f(x) = 1 \;\;\; \forall \; x \in \mathbb{R}
\\ \\
1 = e^{-ix}(\cos x + i \sin x) \Rightarrow e^{ix}=\cos x + i \sin x \;\;\; \forall \; x \in \mathbb{R}
[/math]
>>8621250
>[math] \mathbb { R } [/math]
>>8621250
>[math]\mathbb{R}[/math]
[math] \displaystyle
\\ ax^2 + bx +c = 0 \; \; \; \; \; \; | \; \cdot 4a \\
4a^2x^2 + 4abx + 4ac = 0 \\
4a^2x^2 + 4abx = -4ac \; \; \; \; \; \; | \; +b^2 \\
4a^2x^2 + 4abx +b^2 = b^2 -4ac \\
(2ax + b)^2 = b^2 -4ac \\
2ax + b = \pm \sqrt{b^2 - 4ac} \\
2ax = -b \pm \sqrt{b^2 - 4ac} \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
[/math]