In fact something stronger is true. All of the prime divisors of $(2x)^2+3$ ($x\in\mathbb{N}$) are either equal to $3$ or are congruent to $1$ mod $6$. (And then since they can't all be equal to $3$, there is always some prime divisor congruent to $1$ mod $6$.)
Suppose $p$ is some odd prime dividing $(2x)^2+3$. Note that either $p=3$, or $p\equiv1$ or $p\equiv5$ mod $6$. Assume that $p\equiv5$ mod $6$. Then $$(2x)^2\equiv-3\mod{p}$$ So $-3$ is a square mod $p$. So using Legendre symbols and quadratic reciprocity:$$
\begin{align}
1&=\left(\frac{-3}{p}\right)\\
&=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\\
&=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{(p-1)(3-1)/4}\\
&=\left(\frac{-1}{p}\right)\left(\frac{p}{3}\right)(-1)^{(p-1)/2}\end{align}$$
Since $p\equiv5$ mod $6$, then $p\equiv2$ mod $3$. So $p$ is not a square mod $3$. So continuing:
$$
1=\left(\frac{-1}{p}\right)(-1)^{(p+1)/2}
$$
It is "well-known" that $-1$ is a quadratic residue mod $p$ if and only if $p$ is congruent to $1$ mod $4$. So if $p$ is congruent to $1$ mod $4$, the above says:
$$
1=(1)(-1)
$$
and if $p$ is not congruent to $1$ mod $4$, the above says:
$$
1=(-1)(1)
$$
So it is a contradiction either way. We conclude that all primes dividing $(2x)^2+3$ are either equal to $3$ or are $1$ mod $6$. There is the possibility that $(2x)^2+3$ is a power of $3$, but we can eliminate that easily. Mod $9$, $6$ is not a square. So $(2x)^2+3\not\equiv0$ mod $9$ no matter what $x$ is. So $(2x)^2+3$ is sometimes divisible by $3$ but never divisible by $9$.
So $(2x)^2+3$ is always the product of some primes that are congruent to $1$ mod $6$, sometimes with a single factor of $3$ thrown in the mix. But $(2x)^2+3$ always has prime factors congruent to $1$ mod $6$.
In modulo arithmetic, "division" means multiplying by the multiplicative inverse, e.g., $b = \frac{1}{a}$ means the value which when multiplied by $a$ gives $1$ modulo the value, e.g., $ba \equiv 1 \pmod n$. Note you may sometimes see $b = a^{-1}$ instead to avoid using explicit "division". This works, and gives a unique value, in any cases where the value you're dividing and the modulus are relatively prime.
More generally, it'll work in all cases of $\frac{c}{a} \equiv e \pmod n$ where $d = \gcd(a,n)$ and $d \mid c$ since this gcd value "cancels" in the division. Thus, the resulting equivalent modulo equation of $\frac{c'}{a'} \equiv e \pmod n$, where $c' = \frac{c}{d}$ and $a' = \frac{a}{d}$ has $\gcd(a',n) = 1$, has a solution. However, as Bill Dubuque's comment indicates, this assumes you're doing integer division to the extent of removing the common factor of $d$. Note that $a^{-1}$ doesn't exist modulo $n$ in this case. However, $(a')^{-1}$ does exist modulo $\frac{n}{d}$, so a possible interpretation would be $\frac{c'}{a'} \equiv c'(a')^{-1} \equiv e \pmod{\frac{n}{d}}$.
As for why the multiplicative inverse $b = a^{-1}$ exists modulo $n$ if $\gcd(a,n) = 1$, Bézout's identity states that in such cases there exist integers $x$ and $y$ such that
$$ax + ny = 1 \tag{1}\label{eq1}$$
As such $ax \equiv 1 \pmod n$ so $x \equiv a^{-1} = b \pmod n$. This value must be unique, modulo $n$, because if there was another value $x'$ such that $xa \equiv x'a \equiv 1 \pmod n$, then $(x - x')a \equiv 0 \pmod n$. Since $\gcd(a,n) = 1$, this means that $x - x' \equiv 0 \pmod n \; \iff \; x \equiv x' \pmod n$.
Bézout's identity also shows that if $a$ and $n$ are not relatively prime, e.g., $\gcd(a,n) = d \gt 1$, then \eqref{eq1} becomes
$$ax + ny = d \tag{2}\label{eq2}$$
with the integers of the form $ax + ny$ are always multiples of $d$, so it can't be congruent to $1$ and, thus, $a$ would not have a multiplicative inverse.
Best Answer
$$18x\equiv 1\pmod{25} \\ -7x\equiv1\pmod{25} \\ 7x\equiv-1\pmod{25}\\7x\equiv49\pmod{25} \\ \therefore x\equiv 7\pmod{25}$$