Working out the example from your comment. Here $p=127$, and you want to compute modular square roots of $a=13290059$. First we notice that $a\equiv 17 \pmod p$ and $a\equiv 15892 \pmod {p^2}$. You have apparently found that $12^2\equiv 17 \pmod p$, so the two square roots will be congruent to $\pm12\pmod p$. I do the plus sign here.
We are looking for an integer $b$ such that $b^2\equiv 15892 \pmod{127^2}$ and $b\equiv 12 \pmod {127}$. The latter congruence tells us that $b=12+127 m$ for some integer $m$. Then
$$
b^2=12^2+24\cdot 127 m+127^2m^2\equiv 144+24\cdot127 m\pmod{127^2}.
$$
Here $15892-144=15748=127\cdot124$, so the first congruence simplifies first to
$$144+24\cdot127 m\equiv 15892\pmod{127^2}\Leftrightarrow 24\cdot127 m\equiv 127\cdot124\pmod{127^2},$$
and then by cancelling a factor $127$ to
$$
24m\equiv124\pmod{127}.
$$
The inverse of $24$ modulo $127$ is $90$, so the unique solution of this is
$$
m\equiv 90\cdot124\equiv 111\pmod{127}.
$$
This corresponds to the square root
$$
b=12+111\cdot127=14109.
$$
There are actually several different concepts here, so I'll try to address all of them. I'll get to the modular arithmetic in just a moment, but first a review:
SQUARE ROOTS
We should know that 25 has two square roots in ordinary arithmetic: 5 and -5.
MODULAR ARITHMETIC SQUARE ROOTS
IF the square root exists, there are 2 of them modulo a prime. To continue our example, 25 has the two square roots 5 and -5.
We can check this:
$$(-5)^2 = 25 \equiv 3\bmod 11$$
$$(5)^2 = 25 \equiv 3\bmod 11$$
To find the square roots sometimes takes a bit of trial and error. Often you have to go through each value $v$ and square it (to get $v^2$) to check if it's equivalent to $n \bmod p$, where $n$ is the number whose square root you want to find.
MULTIPLE PRIMES
Again, if a square root exists, there are two square roots modulo each prime. So if we are using multiple primes, there can be more square roots. For example, with two primes, there are 2 square roots modulo the first prime and two square roots modulo the second prime. This gives us $2 \cdot 2 = 4$ square roots.
In general, if we can find a square root modulo each prime, there are a total of $2^n$ square roots modulo $n$ primes.
RETURNING TO YOUR EXAMPLE
We can first calculate 3 modulo 11 and 13:
$$3 \equiv 3 (\bmod 11)$$
$$3 \equiv 3 (\bmod 13)$$
So, modulo 11, we are looking to find a number that, when squared, is equivalent to 3. If we find one, we know that there will be another. So we check the numbers: $1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 9, \dots$ and find that
$$5^2 = 25 \equiv 3 (\bmod 11)$$
...so we know that there will also be another square root modulo 11. Continuing on our quest, we check
$$6^2 = 36 \equiv 3 (\bmod 11)$$
...so we've found the square roots modulo 11. We then continue this modulo 13 to find that:
$$4^2 = 16 \equiv 3 (\bmod 13)$$
$$9^2 = 81 \equiv 3 (\bmod 13)$$
So we know that our square root is either 5 or 6 modulo 11, and either 4 or 9 modulo 13. This gives us 4 possibilities. We can then find that:
$$82 \equiv 5 (\bmod 11), 82 \equiv 4 (\bmod 13)$$
$$126 \equiv 5 (\bmod 11), 126 \equiv 9 (\bmod 13)$$
$$17 \equiv 6 (\bmod 11), 17 \equiv 4 (\bmod 13)$$
$$61 \equiv 6 (\bmod 11), 61 \equiv 9 (\bmod 13)$$
Best Answer
Yes, though it's not well-known (and despite incorrect claims to the contrary in other answers here) it is true that the quadratic formula (completing the square) can be used to solve modular quadratic equations in the way you envision, i.e. by viewing the square-root (and division or fractions) in the formula as multi-valued modular maps. Then - as classically - this allows us to use the formula to reduce solving modular quadratics to "simpler" normalized modular sub-problems of computing a square-root and division (or fraction). I describe this below for modular arithmetic (rings $\Bbb Z/n)$ but readers familiar with ring theory may observe that it works in any commutative ring where $\,2\ \&\ a\,$ are both cancellable (i.e. not zero-divisors), where $\,a\,$ is the lead coefficient.
As a motivating example, below we use this method to correctly do the example in Dan's answer. To make clear how the quadratic formula generalizes modularly, we show the full proof of the quadratic formula by $\color{#0a0}{\text{completing the square}}$ (specialized to this case). As in the OP, we assume that we have available an algorithm to compute modular square roots. Pay close attention to how the modulus changes throughout the process (clarified below).
$$\begin{align} x^2-5x+\,6&\,\equiv\, 0\pmod{\!1000}\\[.1em] \smash{\overset{\times\ 4}\iff}\ 4x^2\!-\!20x\!+\!24&\,\equiv\, 0\pmod{\!4000}\\[.1em] \iff\qquad\ \color{#0a0}{(2x\!-\!5)^2}\!&\,\equiv\, 1\pmod{\!4000}\ \ \ \ \color{#0a0}{\text{complete the }\square}\\[.1em] \iff\qquad\quad\! 2x-5\, &\,\equiv\, \pm\{1,751\}\qquad\ \pmod{\!2000}\\[.1em] \iff\qquad\qquad\quad\ x&\,\equiv\, \color{#c00}{(5\pm\{1,751\})/2}\!\!\!\pmod{\!1000}\\[.1em] \iff\qquad\qquad\quad\ x&\,\equiv\, 2,3,378,-373\ \pmod{\!1000} \end{align}\quad\ $$
We can view this as a quadratic "formula", $ $ i.e. $\, x\equiv \color{#c00}{\dfrac{5\pm\sqrt 1}2}\pmod{\!1000}\ $ iff we correctly view the square root and division maps as $\! $ multi-valued, $ $ and we use the correct modulus throughout. To avoid confusion, it is essential to keep in mind that the "formula" is a concise notation for the result we obtain by $\color{#0a0}{\text{completing the square}}$ - as above.
Note that the modulus $4000$ halves to $2000$ since if $\,r\,$ is a root of $\,x^2\equiv a\pmod{\!4n}\,$ then so too is $\,r\!+\!2n,\,$ by $\,(r\!+\!2n)^2 = r^2\!+\!4n(r\!+\!n)\equiv a\!+\!0.\,$ And it halves again from $2000$ to $1000$ due to the division by $2$, i.e. recall $\,2x\equiv 2a\pmod{\!2n}\iff x\equiv a\pmod{\!n}$.
For a general quadratic $\,ax^2+bx+c\,$ we scale by $\,4a\,$ (vs. $4$ above) when completing the square. So we divide by $2a$ (vs. $2$ above) by using general methods for solving modular linear congruences (or, equivalently, using multi-valued modular fractions - to enable a "formula" view), e.g.
$$\begin{align} 3x^2\,+\,x-4\ &\,\equiv\, 0\ \pmod{\!21}\\[.1em] \smash{\overset{\times\ 12}\iff}\ 36x^2\!+\!12x\!-\!48&\,\equiv\, 0\ \pmod{\!252}\\[.1em] \iff\qquad\ \ \ (6x\!+\!1)^2\! &\,\equiv\, 49\!\pmod{\!252}\\[.1em] \iff\qquad\quad\, 6x\!+\!1\,\ \ &\,\equiv\, \pm7\!\!\!\pmod{\!126}\\[.1em] \iff\qquad\qquad\quad\ \ \ x &\,\equiv\, 1\:\pmod{\!21} \end{align}\qquad\qquad$$
Compared to the more common method of factoring the modulus, then solving the quadratic mod prime powers, then combining the solutions using CRT, this method globally reduces the problem to root-taking and division, vs. that (or other methods) being applied locally mod $p^k$. It may save work by not having to repeat completing the square locally for each prime power, but that is traded off against the fact that there may be more efficient methods of solution after reducing mod $p^k$, e.g. in the prior example our quadratic $f$ reduces $\!\bmod 3\,$ to $\,x\!-\!1,\,$ and $\!\bmod 7\!:\ {-}2f\equiv(x\!-\!1)^2\,$ so the unique root $\,x\equiv 1\pmod{ \!3\ \&\ 7}\,$ lifts to a unique root $\!\bmod 21$ by CCRT (or we could notice the sum of the coef's $= f(1)= 0\,$ so $\,x=1\,$ is a root of $f,\,$ and the cofactor $f/(x\!-\!1) = 3x\!+\!4\!$ yields no more roots mod $3$ or $7)$.