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?

We need $B^2-4AC=0$ that is 
$$1+2c=0 \implies c=-\frac12$$
indeed we obtain 
$$-\frac12(x^2-2x+1)=\left(\frac i{\sqrt2}(x-1)\right)^2$$
Therefore there is no $c$ to have a perfect square on reals.
