Finding the value of $c$ to make $-\dfrac{1}{2}x^2+x+c$ a perfect square trinomial 
I should find the value of $c$ to make: $$-\dfrac{1}{2}x^2+x+c$$ a
  perfect square trinomial.

I really messed up:  $-\dfrac{1}{2}x^2+x+c=x-\dfrac{1}{2}x^2+c$, but it seems like $-\dfrac{1}{2}x^2$ isn't $a^2\boldsymbol{-2ab}+b^2$. Can you help me?

Extract the $-1/2$ from the equation:
$$ -\frac{1}{2}x^2 + x + c = -\frac{1}{2}( x^2 - 2x - 2c )  $$
The expression $ x^2 - 2x - 2c $ is similar to the expression $ x^2-2x + 1 = (x-1)^2 $ which gives that 
$$ -2c = 1 \rightarrow c = -\frac{1}{2} $$
and the final expression is:
$$ -\frac{1}{2}(x-1)^2 $$
