Arithmetic Progresion + Quadratic Equation = Nice Problem

As $c$ , $b$ and $a$ are in arithmetic progression we have $b=c+d$ and $a=c+2d$ . Also $c$ is solution to the quadratic equation, so we have that $(c+2d)c^2+(c+d)c+c=0$ But $c\ne 0$ so: $(c+2d)c+c+d+1=0$ $c^2+(2d+1)c+1+d=0$ But there is only one value for $c$ , so: $(2d+1)^2-4(1+d)=0$ $4d^2=3 \Rightarrow d=\frac{\sqrt{3}}{2}$