How can I find an inverse to the Binet formula?

The second term vanishes quite quickly. We have $F_n=\lfloor \frac{\phi^n}{\sqrt{5}}+\frac{1}{2}\rfloor$, where $\lfloor \cdot \rfloor$ denotes the floor function. If the floor weren't there, you could solve for $n$, and if you rounded you would be wrong only for possibly very small $n$.

Update: A quick calculation shows that $n=\left[ \log_\phi \sqrt{5}(F_n-\frac{1}{2}) \right]$ holds for all $n\ge 3$, and fails for $n=1,2$. In this case $[\cdot]$ denotes the rounding function, or $[x]=\lfloor x+\frac{1}{2}\rfloor$.

I think your formula is not Binet's. Binet's formula is $$F_n=\frac{\varphi^n-\frac{1}{(-\varphi)^n}}{\sqrt{5}},$$ which agrees with your formula for odd $n,$ but not for even $n$. In fact, for even $n$ the numbers your formula gives are not even rational.

So the quadratic formula does not fail. It's just that you've applied it to the wrong formula. Unfortunately the quadratic formula cannot be applied to the correct formula since the correct formula does not contain $\varphi^n$ in both terms—one term contains $(-\varphi)^n$ instead—and so your algebra doesn't go through the same way.

Another remark: the formula you are hoping to find must fail for $n=1$ or $2$ since $F_1=F_2=1$ and you can't have two different numbers as the inverse of $1.$