What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem. 

Determine $x$, such that $3^x+3^{-x}=1$.

This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1±\sqrt{-3}}{2}=0$$
I end up with 
$$\frac{-1±i\sqrt{3}}{2}=0$$
which yields no real solution. And this is not the expected answer.
I'm a 7th grader, by the way. So, I've very limited knowledge on mathematics.
EDIT
I made one interesting observation.
$3^x+3^{-x}$ can be the middle term of a quadratic equation:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$

Just building upon previous comments, it doesn't have as you pointed out a Real result but the Imaginary solution can be analytically found as:
$3^x = \dfrac{1\pm\sqrt{3} i}{2}$
Reexpressing rhs in polar notation:
$3^x = e^{i \dfrac{\pi}{3}}$
And changing lhs basis to $e$
$3^x = e^{\ln{3^x}} = e^{x\ln{3}} $
Then:
$\boxed{x = i  \dfrac{\pi}{3\ln{(3)}}}$
Note: this is the principal value solution. Due to the periodicity of the function, any $x = i  \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$, for $n\in \mathbb{Z}$ will also be a solution. Also the 2nd quadrant values need to be considered $x = - i  \dfrac{\pi}{3\ln{(3)}} + i \dfrac{2\pi n}{\ln{3}}$
