Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$ 
If real numbers $a$, $b$, $c$, $d$ satisfy
  $$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$ 
  then find $(a,b,c,d)$.

What I try: 
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please. 

You can observe this as quadratic equation on $a$ :
$$8a^2-8a(1+b) +(8b^2+8c^2  -8bc+7)=0$$ which has to have a solution so the discriminant must be nonnegative:
$$64(1+b)^2-32(8b^2+8c^2  -8bc+7)\geq 0$$
so $$6b^2-4b(1+2c)+8c^2+5\leq 0$$
Now again, the discriminat must $\geq 0$ since that inequality has a solution... 
