19. Let $A = \{ f(x) \mid x \in \mathbb{R} \}$ and $f(0) = c$. Plugging in $x = y = 0$ we get $f(-c) = f(c) + c - 1$, hence $c \neq 0$. If $x \in A$, then taking $x = f(y)$ in the original functional equation we get $f(x) = \frac{c+1}{2} - \frac{x^2}{2}$ for all $x \in A$. We now show that $A - A = \{ x_1 - x_2 \mid x_1, x_2 \in A \} = \mathbb{R}$. Indeed, plugging in $y = 0$ into the original equation gives us $f(x - c) - f(x) = cx + f(c) - 1$, an expression that evidently spans all the real numbers. Thus, each $x$ can be represented as $x = x_1 - x_2$, where $x_1, x_2 \in A$. Plugging $x = x_1$ and $f(y) = x_2$ into the original equation gives us $f(x) = f(x_1 - x_2) = f(x_1) + x_1 x_2 + f(x_2) - 1 = c - \frac{x_1^2 + x_2^2}{2} + x_1 x_2 = c - \frac{x^2}{2}$. Hence we must have $c = \frac{c+1}{2}$, which gives us $c = 1$. Thus $f(x) = 1 - \frac{x^2}{2}$ for all $x \in \mathbb{R}$. It is easily checked that this function satisfies the original functional equation.