If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . . 
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that
  $$a^2 + b^2 + c^2=d$$
  then $a+b+c+d$ equals?
  $$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$
  Here is what I tried: assume the common difference is $x$.
  Thus,
  $$a^2 + a^2 + 2ax + x^2 +a^2 +4ax +4x^2=a + 3x$$
  That is,
  $$3a^2 + 6ax + 2x^2=a + 3x$$
  First I assumed it as a quadratic in $a$ and solved for it.
  $$a=\frac{(1-6x) \pm \sqrt{12x^2-48x +1}}{6}$$ but to no avail.
  Then I assumed it as a quadratic in $x$ and got:
  $$x=\frac{(3-6a) \pm \sqrt{12a^2 -36a +81}}{6}$$
  And I noticed that I'm stuck. Please help. Thanks in advance!

Edit: I made a calculation error as pointed out by multiple answers, and so the equation should have been $$3a^2 + 6ax + 5x^2=a + 3x$$ which leads to different solutions.

Let $x$ be the common difference.

From $a^2+b^2+c^2=d$, we get $d > 0$,

Since $a,b,c,d$ are distinct, we get $x\ne 0$.

If $x < 0$, then $c > d$, hence 
$$c^2 > d^2\ge d=a^2+b^2+c^2 > c^2$$
contradiction.

Hence $x > 0$.

\begin{align*}
\text{Then}\;\;&a^2+b^2+c^2=d\\[4pt]
\implies\;&(b-x)^2+b^2+(b+x)^2=b+2x\\[4pt]
\implies\;&3b^2+2x^2=b+2x\\[4pt]
\implies\;&b(1-3b)=2x(x-1)\\[4pt]
\implies\;&b(1-3b)\ge 0\\[4pt]
\implies\;&b=0\;\;\\[4pt]
\implies\;&2x(x-1)=0\\[4pt]
\implies\;&x=1\\[4pt]
\implies\;&a,b,c,d=-1,0,1,2\\[4pt]
\implies\;&a+b+c+d=2\\[4pt]
\end{align*}
