Solve the function equation $g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0$ let $g(x)\in \Bbb R$ and for any $x\in \Bbb R$ such that
$$g^2(x)-g(x+1)-\dfrac{x^2+2x-6}{4}=0,   g(0)=0$$ 
find $g(x)$
my idea
let $x\longrightarrow x+1$, then we have
$$g^2(x+1)-g(x+2)-\dfrac{(x+1)^2+2(x+1)-6}{4}=0$$
$$g^2(x+2)-g(x+3)-\dfrac{(x+2)^2+2(x+2)-6}{4}=0$$
$$\cdots\cdots $$
$$g^2(x+n)-g(x+n+1)-\dfrac{(x+n)^2+2(x+n)-6}{4}=0$$
following  I can't work,Thank you 

Take $\,x=-1\;$:
$$g(-1)^2-g(0)-\frac{(-1)^2+2(-1)-6}4=0\implies g(-1)^2=-\frac74$$
which of course is impossible unless there are further conditions or we're working with complex numbers and not with the reals, as written, say.
