I tried to use the definition: $$\displaystyle \Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-{s^2}/{2}}\,\mathrm ds$$
So, according to this site: $$\int \:e^{-{x^2}/{2}}\mathrm dx=\frac{\sqrt{\pi }}{\sqrt{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right)+C$$
But by definition: $${\displaystyle \operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-s^{2}}\,\mathrm {d} s}$$
I do not know how to follow after the function $ erf (...) $
Maybe the value is only possible to get it through tables?
How to determine $x,x∈R^+$ such that $\Phi(x)=0,9505$?
Thank you very much.
If you use the definitions, you end with $$\displaystyle \Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-{s^2}/{2}}\, ds=\frac{1}{2} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$ So, if you need to solve for $x$ equation $\Phi(x) =c$, the solution is $$x=\sqrt{2}\, \text{erf}^{-1}(2 c-1)$$
If you do not want to use tables, you could use the expansions given here
$$\text{erf}^{-1}(z)=\frac{1}{2}\sqrt{\pi}\left (z+\frac{\pi}{12}z^3+\frac{7\pi^2}{480}z^5+\frac{127\pi^3}{40320}z^7+\frac{4369\pi^4}{5806080}z^9+\frac{34807\pi^5}{182476800}z^{11}+\cdots\right)$$ The problem is that this expansion is very slowly convergent when the argument is large as in your case $(2c-1=0.901)$; the table below gives some results as a function of the number of terms used $$\left( \begin{array}{cc} n & x \\ 1 & 1.129 \\ 2 & 1.369 \\ 3 & 1.476 \\ 4 & 1.535 \\ 5 & 1.571 \\ 6 & 1.595 \\ 7 & 1.610 \\ 8 & 1.621 \\ 9 & 1.629 \\ 10 & 1.634 \\ 11 & 1.638 \\ 12 & 1.641 \\ 13 & 1.643 \\ 14 & 1.645 \\ 15 & 1.646 \\ 16 & 1.647 \\ 17 & 1.647 \\ 18 & 1.648 \\ 19 & 1.648 \\ 20 & 1.649 \\ 21 & 1.649 \end{array} \right)$$
You could also use the approximation $$\operatorname{erf}(z) \approx \sqrt{1 - \exp\left(-z^2\frac{\frac{4}{\pi} + az^2}{1 + az^2}\right)}$$ given in the Wikipedia page which then reduce to a quadratic equation in $z^2$ (use $a=0.147$).
For the vale you give, this would lead to $x=1.64953$ while the exact value is $x=1.64972$.