From the first one,
expanding the terms,
$\begin{array}\\
931
&=36(n + (d+6)/12)^2 - 24(n + (d+6)/12) - (d^2 -16)/4\\
&=36(n^2+n(d+6)/6 + (d+6)^2/144) - 24n -2 (d+6)) - d^2/4 +4\\
&=36n^2+6n(d+6) + (d+6)^2/4 - 24n -2 (d+6) - d^2/4 +4\\
&=36n^2+6nd+36n + (d^2+12d+36)/4 - 24n -2 d-12 - d^2/4 +4\\
&=36n^2+6nd+36n + d^2/4+3d+9 - 24n -2 d-12 - d^2/4 +4\\
&=36n^2+6nd+12n +d+1\\
\end{array}
$
But this is always
one more that
$36n² + 12n +6dn +d
$,
so the second equation
does not provide
any additional information.
Note that
$(6n+d/2)^2
=36n^2+6nd+d^2/4
$,
so your expression equals
$\begin{array}\\
6n² +6dn+d^2/4-d^2/4+ 12n +d+1
&=(6n+d/2)^2-d^2/4+ 2(6n +d/2)+1\\
&=((6n+d/2)+1)^2-d^2/4\\
\end{array}
$
If $d = 2c$,
then
$931
=(6n+c+1)^2-c^2
$
or
$\begin{array}\\
931
&=7^2 19\\
&=(6n+c+1)^2-c^2\\
&=(6n+c+1-c)(6n+c+1+c)\\
&=(6n+1)(6n+2c+1)\\
&=(6n+1)(6n+d+1)\\
\end{array}
$
If $n$ and $d$ are integers
then
$n = 3$ and $d = 30$
and there are
no other solutions in
positive integers
(since the only possible
values for
$6n+1$
are
$19$ and $49$).
If you don't care
about $n$ and $d$ being integers,
you can choose $n$ arbitrarily
and let
$d
=\frac{931}{6n+1}-(6n+1)
$.
(added later)
More generally,
if we want the value
to be $m$
instead of $931$,
we can get the following
sequence of restrictions
on $n$ and $d$.
First of all,
we want
$m
=(6n+1)(6n+d+1)
$.
So,
if there are
no restrictions on
$n$ and $d$,
choose $n$ arbitrarily
and let
$d
=\frac{m}{6n+1}-(6n+1)
$.
If $n \ge 0$ and
$d \ge 0$,
then
$m
\ge (6n+1)^2
$
or
$n
\le \dfrac{\sqrt{m}-1}{6}
$.
If, in addition,
we want $n$ and $d$
to be integers,
then
$(6n+1) | m$
and
$d
=\frac{m}{6n+1}-(6n+1)
$.
