Proposition: $\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = \frac{1 + \sqrt{1 + 4x}}{2}$.

First, I would like to mention that said series is not necessarily well-defined; you can't just add a "..." and expect the resulting quantity to be well-defined. For example, consider: $$S=1+2+4+8+16+\ldots$$ Then, $$2S=2+4+8+16+\ldots$$ so: $$2S+1=S$$ and you get $S=-1$, which is clearly absurd. The issue here is that $S$ is not well defined; you have to define $S$ as the limit as $n \rightarrow +\infty$ of $1+2+\ldots+2^n$, and one can show that this quantity is $+\infty$. Similarly, you must instead ask for the limit of the sequence $a_n$, where $a_0=\sqrt{x}$ and $a_{n+1}=\sqrt{x+a_n}$. Now one can show that this sequence is increasing and is upper bounded, so it must converge (for $x>0$). Only once you have shown that it has converged can you use Nemo's approach (the comments about the quadratic formula's two solutions are easily resolved since the limit has to be positive).

it is true because : $$S=\sqrt { x+\sqrt { x+\sqrt { x+... } } } \\ S=\sqrt { x+S } \\ { S }^{ 2 }-S-x=0\\ S=\frac { 1+\sqrt { 1+4x } }{ 2 } $$