For natural numbers, $a,b,c$ the Gauss sum is defined as
$$G(a,b,c)=\sum_n^{c-1}e^{2\pi i\frac{an^2+bn}{c}}$$
From the quadratic Gauss sum Wikipedia page it is given that for odd $c$ and $\text{gcd}(a,c)=1$, one has
$${\displaystyle G(a,b,c)=\varepsilon _{c}{\sqrt {c}}\cdot \left({\frac {a}{c}}\right)e^{-2\pi i{\frac {\psi (a)b^{2}}{c}}}.}$$
where $\psi(a)$ is some number with $4\psi(a)a ≡ 1 \mod c$ and
$$\varepsilon_{m}=\begin{cases}1&\text{if}\ m\equiv 1\pmod 4,\\ i&\text{if}\ m\equiv 3\pmod 4\end{cases}$$
Note that for $b=0$ and odd $c$ it is given that
$$G(a,b=0,c)=\varepsilon_c\sqrt c \left(\frac a c\right)$$
so there is at least an implicit step which allows us to write
$${\displaystyle G(a,b,c)=G(a,0,c)e^{-2\pi i{\frac {\psi (a)b^{2}}{c}}}.}$$
But completing the square will result in a summation which looks like
$$G(a,b,c)=\sum_n^{c-1}e^{2\pi i\frac{a(n+b/2a)^2}{c}}e^{-2\pi i\frac{a(b/2a)^2}{c}}$$.
What is the missing step which allows us to ignore the $b/2a$ in the quadratic term?