I do not think that any closed form could be obtained for the solution of $$f(t)=\sqrt{\pi } t \left(\text{erf}\left(\frac{t}{\sqrt{2}}\right)+1\right)-\sqrt{2}
\alpha e^{-\frac{t^2}{2}}=0$$ and I suppose that numerical methods should be required. Admitting that Newton method would be your choice, you need $$f'(t)=\sqrt{\pi } \left(\text{erf}\left(\frac{t}{\sqrt{2}}\right)+1\right)+\sqrt{2}
(\alpha +1) e^{-\frac{t^2}{2}} t$$ and the iterates will be given by $$t_{n+1}=\frac{\sqrt{2} \left(\alpha +(\alpha +1) t_n^2\right)}{\sqrt{\pi } e^{\frac{t_n^2}{2}}
\left(\text{erf}\left(\frac{t_n}{\sqrt{2}}\right)+1\right)+\sqrt{2} (\alpha +1) t_n}$$ But, as usual, the problem is to find a "reasonable" guess of the solution.
If $t$ is supposed to be small, you could use Taylor expansion $$f(t)=-\sqrt{2} \alpha +\sqrt{\pi } t+\left(\frac{\alpha }{\sqrt{2}}+\sqrt{2}\right)
t^2+O\left(t^4\right)$$ and then start using either $$t_0=\sqrt{\frac{2}{\pi }} \alpha$$ if you limit the expansion to $O\left(t^2\right)$ or $$t_0=\frac{\sqrt{4 \alpha (\alpha +2)+\pi }-\sqrt{\pi }}{\sqrt{2} (\alpha +2)}$$ if you limit the expansion to $O\left(t^4\right)$.
Because of the exponential term, assuming $\alpha>0$, I would prefer to solve $$g(t)=\log \left(\sqrt{\pi } t
\left(\text{erf}\left(\frac{t}{\sqrt{2}}\right)+1\right)\right)-\log
\left(\sqrt{2} \alpha e^{-\frac{t^2}{2}}\right)=0$$ $$g'(t)=\frac{\sqrt{\frac{2}{\pi }}
e^{-\frac{t^2}{2}}}{\text{erf}\left(\frac{t}{\sqrt{2}}\right)+1}+t+\frac{1}{t}$$
Edit
For $0\leq t\leq 1$ (this corresponds to $0 \leq \alpha\leq 3.5$), it seems that a very good approximation could be $$t\approx -0.019727 \sqrt{\alpha}+0.992234 \alpha-0.575731 \alpha^{3/2}+0.109344 \alpha^2\qquad (R^2=0.999999)$$ For example, using $\alpha=2$, the above correlation gives $t\approx 0.765533$ for an "exact" solution $t=0.765277$.
Added after @tired's comment
As @tired commented, if looking for large values of $\alpha$, the solution is almost the one of
$$h(t)=2\sqrt{\pi } t -\sqrt{2}
\alpha e^{-\frac{t^2}{2}}=0\implies t^2=W\left(\frac{\alpha ^2}{2 \pi }\right)$$ where $W(z)$ is Lambert function. For large values of $\alpha$ , we can use as a good approximation
$$t^2=L_1-L_2+\frac {L_2}{L_1}\qquad L_1=\log \left(\frac{\alpha ^2}{2 \pi }\right)\qquad L_2=\log(L_1)$$ Trying $\alpha=123.456$, the above asymptotics leads to $t=2.45031$ while the "exact" solution would be $t=2.45235$.
