I had to solve this step by step, but teacher said that my solution is wrong, so I didn't get full marks. My solution is this:
$\frac{16}{2} - 4^{2} - (-3)^{2}$ = $\frac{16}{2}- 4^{2} - 9 $= $\frac{16}{2} - 16 - 9 $= $8 - 16 - 9 $= $-8 - 9$ = $-17$
His solution is this:
$\frac{16}{2} - 4^{2} - (-3)^{2}$ = $\frac{16}{2} - 4^{2} - (9)$ = $\frac{16}{2}- 16 - 9 $= $8 - 16 - 9 $= $-8 - 9$= $-17$
Is he right about it or should I complain?
If the only difference between one line and the other is the parenthesis then yes, you should complain. Leaving the parenthesis around the $9$ after you squared the $3$ adds no value whatsoever to the computation of that expression.