Modular arithmetic and inverse functions The question is shown below:
Suppose $S=\{0,1,2,3,4,5,6,7,8,9,10\}$ and that the function $f:S\rightarrow S$ is given by:
$f(x)=6x^2+3x+8$ (mod 11)
Let $T=\{0,5\}$.
Find $f^{-1}\left(T\right)$.
Alright,
so my initial approach with this question was to find the inverse function,
$f^{-1}(x)=\dfrac{\pm\sqrt{24x-183}-3}{12}$, but $f^{-1}x\in S\geq0$
thus,
$f^{-1}(x)=\dfrac{\sqrt{24x-183}-3}{12}$
and then just churn out congruent values of $x=0,5$ in modulo 11 until I find a suitable value for $f^{-1}(x)$ within $S$, but it seems a bit tedious. Anyways, I repeated this churning process until I received the values of $\{2,3,6,10\}$.
Can anyone suggest a faster approach and please advise me if my approach is logically sound.
Many thanks :)

If $\ a\equiv \alpha^2\pmod{11}\ $ is a quadratic residue $\mod{11}\ $, then
$$
a^6 \equiv \alpha^{12}\equiv \alpha^2\equiv a\pmod{11}\ ,
$$
so the square roots of $\ a\pmod{11}\ $ are $\ \pm a^3\pmod{11}\ $.
If $\ y\equiv 6x^2+3x+8\pmod{11}\ $ then $\ 2y+4\equiv$$ x^2+6x+9$$\equiv(x+3)^2 \pmod{11}\ $ must be a quadratic residue, so $\ (2y+4)^3\equiv$$\pm(x+3) \pmod{11}\ $, and $\ x\equiv$$\pm(2y+4)^3-3\pmod{11}\ $.
Substituing $\ y=0\ $ in this expression gives $\ x\equiv$$ \pm4^3-3\equiv$$6,10\pmod{11}\ $. Substituting one of $\ x=6,10\ $ back into the polynomial $\ 6x^2+3x+8\ $ shows that it is indeed congruent to $\ 0\mod11\ $ for both of them (this check is necessary for at least one of these two values, to eliminate the possibility that $\ 2\cdot5+4\ $ might have been a quadratic non-residue $\mod 11\ $).
Similarly, substituting $\ y=5\ $ into the expression for $\ x\ $ gives $\ x\equiv$$\pm3^3-3\equiv$$2,3\pmod{11}\ $, which again can be shown to satisfy the congruence $\ 5\equiv 6x^2+3x+8\pmod{11}\ $. Thus, $\ f^{-1}\left(\{0,5\}\right)=$$\{2,3,6,10\}\ $, as you've already established.
