(2c^2+9)-(3c^2-7)
$\displaystyle (2c^2+9)-(3c^2-7) $
Multiply through the 2nd bracketed term by the negative!
$\displaystyle (2c^2+9)-3c^2-(-7) $
$\displaystyle 2c^2+9-3c^2+7 $
Collect like terms:
$\displaystyle -c^2+16 $
Change the order so the positive term is first!
$\displaystyle 16-c^2 $
Recognise that this is a difference of two squares since $\displaystyle 16 = 4^2 $:
$\displaystyle 4^2-c^2 $
$\displaystyle (4-c)(4+c) $