jmc

We square both sides of the equation to get $$\begin{array}{rl}\sqrt{x+1}& =x\\ x+1& =x^2\\ x^2-x-1& =0\end{array}$$We can solve for $x$ either by completing the square or applying the quadratic formula, which gives us the smaller solution $x=\frac{1-\sqrt{5}}{2}$ and the larger solution $x=\frac{1+\sqrt{5}}{2}$. Consequently, $a=1$, $b=5$, and $c=2$, so $a+b+c=\boxed{8}$.

Note that the larger of these two roots is just the value of $\varphi$, the golden ratio.