So many square roots!

Let

$\begin{aligned} y & = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}} & \small \blue{\text{Squaring both sides}} \\ y^2 & = x + \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}} \\ & = x + y \\ \implies x & = y^2 - y = y(y-1) \end{aligned}$

For $y=2019 \implies x = 2019\times 2018 = \boxed{4074342}$ .

Let the L.H.S. of the equation be $y$ . Then $y=\dfrac{1+\sqrt {4x+1}}{2}$ (since $y$ is positive). Solving for $x$ , we get $x=y(y-1)$ . Here $y=2019$ . So $x=2019\times {2018}=\boxed {4074342}$

Bonus : The given equation has real solution when $x$ is non-negative definite, which is true when R.H.S. is greater than or equal to $\boxed {1}$

$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{...}}}}=2019$

Squaring both sides gives

$x+\sqrt{x+\sqrt{x+\sqrt{...}}}=2019^2$

However, $\sqrt{x+\sqrt{x+\sqrt{...}}}=2019$

So

$x+2019=2019^2$

$x=2019^2-2019$

$x=2019(2019-1)$

$x=2019(2018)$

$x=\color{#69047E}\large{\boxed{4074342}}$