Very quadratic

The equation has a solution if the discriminant is non-negative. We set $A = c,\ \ B = -2c,\ \ C = 6-c^2$ so that $D = B^2 - 4AC = (2c)^2 - 4\cdot c\cdot (6-c^2) = 4c^2 - 24c + 4c^3 \geq 0.$ To find the zeroes of $D$ , we factor $D = 4c(c^2 + c - 6) = 4c(c+3)(c-2).$ For sufficiently small $c$ , the three factors are negative, so that $D < 0$ . The smallest value where this is no longer the case is $c = \boxed{-3}$ .

Here is a graph of the discriminant $D$ as a function of the parameter $c$ :

And this graph shows the solutions of $f_c(x) = 0$ as a function of $c$ :

We can write $f_c(x)=c(x-1)^2-c^2-c+6$ . For negative $c$ , the equation $f_c(x)=0$ has a real solution iff $c^2+c-6= (c+3)(c-2)\leq 0$ . The smallest such $c$ is $c=\boxed{-3}$ .