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. 

Calculalte discriminant twice and the discriminat tells the value is only one. 
$2a^2-2(b+1)a+b^2+(b-c)^2+c^2+1-1/4=0 $
$D/4=(b+1)^2-2{2b^2-2bc+2c^2+3/4}\geq0$
We get,
$3b^2-2(1+2c)b+4c^2+1/2=0 $
$D/4=(1+2c)^2-3(4c^2+1/2)\geq0 $
$-8c^2+4c-1/2\geq0 $
$16(c-1/4)^2\leq0 $
$c=1/4 $
$3b^2-3b+3/4<=0 $
$(2b-1)^2\leq0 $
$b=1/2$ 
$2a^2-3a+1/4+1/16+1/16+1-1/4=0 $
$16a^2-24a+9=0 $
$(4a-3)^2=0$
$a=3/4$
