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Is it possible to get the right formula for vega of a call option under the black scholes model from this formula?
$$\frac{\partial{C}}{\partial{\sigma}}=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)$$
$d_-=\frac{\ln{\frac{S_0}{k}}+(r-\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$ $d_+=\frac{\ln{\frac{S_0}{k}}+(r+\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$
Note that, \begin{align*} \frac{\partial{C}}{\partial{\sigma}} &=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)\\ &=\frac{1}{\sqrt{2\pi}}e^{\frac{-d_+^2}{2}}\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{d_+^2}{2} - \frac{d_-^2}{2}} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{1}{2}(d_+-d_-)(d_++d_-)} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{Ke^{-rt}d_+}{\sigma} e^{\frac{1}{2}\sigma \sqrt{t}\, \frac{2\ln \frac{S_0}{K} +2rt}{\sigma \sqrt{t}}} \right]\\ &=N'(d_+)\left[-\frac{S_0 d_-}{\sigma} + \frac{S_0d_+}{\sigma} \right]\\ &=S_0 N'(d_+)\sqrt{t}, \end{align*} which is the Black-Scholes vega formula.
The answer by @Gordon is pretty complete, but let me add one more point. Let $n(x) = N'(x)$ be the PDF of standard normal distribution.
In the derivation, note that $$ e^{d_+^2/2 - d_-^2/2} = \frac{n(d_-)}{n(d_+)} = \frac{S_0}{Ke^{-rt}}. $$ Thanks to this relation, there are two equivalent expressions for the Black-Scholes vega: $$ \frac{\partial C}{\partial \sigma} = S_0 n(d_+) \sqrt{t} = K e^{-rt} n(d_-) \sqrt{t}. $$ See Wikipedia.
