how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$ I’d like to solve following equation for $x.$
If it is not possible, why I can’t?
$$\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$$
 A: There always exists at least one solution: If $x >\mu_i$ for all $i$, then each of the terms in the sum is positive and if $x < \mu_i$ for all $i$, then each of the terms in the sum is negative.  Thus, the function on the left hand side has at least one zero and all the zeroes lie strictly between the lowest of the $\mu_i$ and the highest.  
Finding an explicit analytic solution is almost certainly not possible, but numerically, it will be easy.  For example, starting with the smallest interval that contains the $\mu_i$, use the bisection method, which will converge exponentially fast to at least one root.
A: Here is why solutions do exist. $\newcommand{\bR}{\mathbb{R}}$ Consider the function $f:\bR\to\bR$
$$
f(x)=\sum_{i=1}^n e^{-(x-\mu_i)^2}.
$$
Note that 
$$
0\leq f(x)\leq n, \;\;\forall x\in\bR
$$
and
$$
\lim_{\vert x\vert\to\infty} f(x)= 0.
\tag{$1$}
$$
Set 
$$
M:=\sup_{x\in \bR} f(x).
$$
If $(x_\nu)$ is a sequence of real numbers such that
$$
\lim_{\nu\to\infty }f(x_\nu)=M,
$$
then, due to (1), the sequence is bounded and thus admits a  convergent subsequence. The limit $x^*$ of this subsequence satisfies $f(x^*) =M$ so
$$
f'(x^*)=0.
$$ 
The last equation is equivalent to yours
$$
\sum_{i=1}^n (x^*-\mu_i) e^{-(x^*-\mu_i)^2}=0.
$$
Using more sophisticated mathematics (namely $o$-minimality) one can say a bit more.  $\newcommand{\eZ}{\mathscr{Z}}$
Consider the set
$$\eZ:=\Big\{\; (x,\mu_1,\dotsc, \mu_n)\in\bR^{n+1}:\;\;\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2}=0\;\Big\}.
$$
We have a natural projection 
$$
\pi:\eZ\to\bR^n,\;\;\eZ\ni (x,\mu_1,\dotsc,\mu_n)\mapsto (\mu_1,\dotsc,\mu_n).
$$
For $\vec{\mu}\in\bR^n$  we set
$$
\eZ(\vec{\mu}):=\pi^{-1}(\vec{\mu}).
$$
The above discussion shows that $\eZ(\vec{\mu})$ is a nonempty discrete   subset of $\bR$. A famous result of Alex Wilkie 

Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function, J Amer Math Soc, 9, (4), 1996, 1051-1094.

implies that there exists a natural number $N$ such that
$$
0<\# \eZ(\vec{\mu})\leq N,\;\;\forall \vec{\mu}\in\bR^n.
$$
A more precise  statement is true. There exist  finitely many pairwise disjoint open subsets $\Omega_1,\dotsc, \Omega_\nu$ of the $\mu$ space  with the following properties


*

*The union of the closures of the $\Omega_i$-s  is the entire $\mu$ space.

*The boundaries of $\Omega_i$-s are piecewise $C^2$-hypersurfaces. (I won't elaborate on the precise meaning of "piecewise $C^2$".)

*The function $\mu\mapsto \#\eZ(\mu)$ is constant on each $\Omega_i$.


As others have indicated, your equation can have multiple solutions. E.g., for $n=5$ and $\mu_i=2i$ the graph of
$$
g(x)=\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2}
$$ 
is depicted below (courtesy of MAPLE).

As you can see, the graph of $g(x)$ intersects the $x$-axis in $8$ points.
Consider a point $(x_0,\mu)\in\eZ$ which is a regular point of the projection $\pi:\eZ\to\bR^n$.  In other words
$$
\partial_x G(x_0,\vec{\mu})\neq 0,\;\;G(x,\vec{\mu})=\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2}.
$$
The implicit function theorem  shows  that near $(x_0,\vec{\mu})$ the solution $(x,\vec{\mu})$ of $G(x,\vec{\mu})$   depends smoothly on $\vec{\mu}$. Moreover $\newcommand{\pa}{\partial}$
$$ 
\frac{\pa G}{\pa x}\frac{\pa x}{\pa \mu_i}+\frac{\pa G}{\pa\mu_i}=0
$$
so $$
\frac{\pa x}{\pa\mu_i}=-\frac{G'_{\mu_i}}{G'_x}
$$
We have
$$G'_{\mu_i}= e^{-(x_0-\mu_i)^2}\Big(\; 2(x_0-\mu_i)^2-1\;\Big)
$$
$$
G'_x= \sum_{j=1}^n e^{-(x_0-\mu_i)^2}\Big(1-2(x_0-\mu_j)^2\Big).
$$
We deduce that
$$
\sum_i \frac{\pa x}{\pa\mu_i}=1.
$$
In particular
$$
\frac{d}{dt}x(\mu_1+t,\dotsc, \mu_n+t)=1
$$
so
$$
x(\mu_1+t, \dotsc, \mu_n+t)=x(\mu_1,\dotsc, \mu_n)+t.
$$
Thus, as observed by Neil Strickland in his answer, it suffices  to determine  $x(\mu_1,\dotsc,\mu_n)$ only when $\sum_i\mu_i=0$.
`
A: It is simple to reduce to the case where $\sum_i\mu_i=0$.  Then the simplest nontrivial case is $(x-\mu)e^{-(x-\mu)^2}+(x+\mu)e^{-(x+\mu)^2}=0$, which is equivalent to $(x+\mu)/(x-\mu)=e^{4x\mu}$.  Here there is a solution of the form
$$ x = \frac{1}{\sqrt{2}} \left(1 + \frac{1}{3}\mu^2 + \frac{11}{90}\mu^4 + \frac{17}{630}\mu^6 - \frac{281}{37800}\mu^8 - \frac{44029}{3742200}\mu^{10} + O(\mu^{12}) \right)$$ 
I have calculated further terms as far as $\mu^{240}$, and there is some rather interesting behaviour, with some features of the coefficient of $\mu^i$ apparently depending on $i$ mod $14$; I don't know how to explain that.  The radius of convergence appears to be finite, perhaps close to $1.1$, but the evidence for that is not hugely compelling.  Note that $281$ and $44029$ are prime; this means that various classes of simple formulae cannot produce the above list of coefficients.  (The coefficient of $\mu^{12}$ is $-12147139/2043241200$, which also has prime numerator, but this pattern breaks down for $\mu^{14}$.)  
A: Since the question is how to solve the equation, you can look at the bisection method, which is a recursive application of the idea in @Rober answer.
