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

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

Do you mean:

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