$1200[{[1 + (.07/12) ] 180 – 1} / (.07/12)] =

$1200[{[1 + (.07/12) ] 180 – 1} / (.07/12)] = ?

Do you mean:

$$\small{\text{
$r = \frac{7}{12}\ \% $
and after
$n = 180 $
months, the sum
$
s = \$ \ 1200 * \left( \dfrac{ (1+r)^n - 1 } { r } \right)
$
}}\\
\small{\text{
$
s = \$ \ 1200 * \left( \dfrac{ \left( 1+ \dfrac{0.07}{12} \right)^{180} - 1 } { \dfrac{0.07}{12} } \right)
= \$ \ 1200 * \left( \dfrac{ 2.84894673087 - 1 }{ 0.00583333333 }\right)
$
}}\\\\
\small{\text{
$
= \$ \ 1200 * 316.962296721 = \$ \ 380354.76
$
}}$$