How to solve this quadratic congruence equation How to solve $x^2+x+1\equiv 0\pmod {11}$ ?
I know that in some equations like $ax\equiv b\pmod d$ if $(a,d)=1$ then the equation has one and only one solution $x\equiv ba^{\phi(d)-1}\pmod d$. Any help will be appreciated. ;)

You can use the technique of Completing the Square.
Suppose that we want to solve the congruence $ax^2+bx+c\equiv 0 \pmod p$, where $a\not\equiv 0\pmod{p}$, and $p$ is an odd prime. The congruence is equivalent to
$$4a^2x^2+4abx+4ac\equiv 0\pmod{p}.\tag{$1$}$$
Completing the square, we see that the congruence is equivalent to 
$$(2ax+b)^2-b^2+4ac\equiv 0\pmod{p},$$
or equivalently to
$$(2ax+b)^2\equiv b^2-4ac\pmod{p}.\tag{$2$}$$
To solve this last congruence we (i) Try to solve the congruence $y^2 \equiv b^2-4ac\pmod{p}$.  There may be no solution, in which case the original congruence has no solution. There may be a single solution, if $b^2-4ac\equiv 0\pmod{p}$. In all other cases there will be two solutions. (ii) After having found such $y$, we solve the linear congruences $2ax+b\equiv y\pmod{p}$.
Now we turn to $x^2+x+1\equiv 0\pmod{p}$, and use the basic technique, although inspection would work faster.
Rewrite as $4x^2+4x+4\equiv 0\pmod{11}$, then as $(2x+1)^2-1+4\equiv 0\pmod{11}$, then as $(2x+1)^2\equiv -3\equiv 8\pmod{11}$.
So we first try to solve the congruence $y^2\equiv -3\equiv 8\pmod{11}$. 
It turns out there are no solutions. There are powerful methods to deal with such questions for large primes $p$. For $11$, just calculate $1^2$, $2^2$, $3^3=2$, $4^2$, and $5^2$ modulo $11$. None of them work.  The rest cannot work, for they are essentially the negatives of the numbers we have already checked, so their squares are, modulo $11$, the same as the squares of the numbers $1$ to $5$.  For example, $6\equiv -5\pmod{11}$, so $6^2\equiv (-5)^2\equiv 5^2\pmod{11}$, and similarly $7^2\equiv 4^2$, and so on.
Remarks: $1.$ The congruence $(2)$ has a solution if and only if the congruence $y^2\equiv b^2-4ac\pmod{p}$ has a solution, that is, iff $b^2-4ac$ is a square modulo $p$. This $b^2-4ac$ is the familiar discriminant of elementary algebra. 
$2.$ The general idea can be used even when we are solving a quadratic congruence modulo $m$, where $m$ is not prime.  But significant modifications need to be made. The congruence $ax^2+bx+c\equiv 0\pmod{m}$ is equivalent to $4a^2x^2+4abx+4ac\equiv 0 \pmod{4am}$, so we end up looking at 
$$(2ax+b)^2\equiv b^2-4ac\pmod{4am}.$$
The study of the congruence $y^2\equiv b^2-4ac\pmod{4am}$ is more complicated than when the modulus is $p$, and there may be many solutions.  But after that we are solving linear congruences $2ax+b\equiv y\pmod{4am}$, and we can use the Extended Euclidean Algorithm. 
