Mistake in solving an equation involving a square root I want to solve $2x = \sqrt{x+3}$, which I have tried as below:
$$\begin{equation}
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\left(x - \frac12 \right)^2 = 1 \\
x = \frac32 , -\frac12
\end{equation}$$
This, however, is incorrect.
What is wrong with my solution?

$\require{cancel}$Additional details in $\color{blue}{blue}$.  Important detail in $\color{green}{green}$.  Mistakes in $\color{red}{\cancel{\text{canceled red}}}$. Corrections in $\color{purple}{purple}$.
$\begin{equation}
\color{blue}{2x = \sqrt{x+3}}\color{green}{\ge 0}\\
\color{blue}{4x^2 = x+3}\color{green}{\text{!AND!} x\ge 0}\\
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\color{blue}{x^2 -2\cdot\frac 18=\frac 34}\\
\color{blue}{x^2 -2\cdot\frac 18 +(\frac 18)^2=\frac 34+\frac 1{64}}\\
\color{red}{\cancel{(x - \frac12 )^2 = 1}}\color{purple}{(x - \frac18)^2 = \frac {49}{64}} \\
\color{blue}{x-\frac 18=\pm \frac 78}\\
\color{red}{\cancel{x = \frac32 , -\frac12}}\color{purple}{x = 1 , -\frac34}\color{green}{\text{!AND!} x\ge 0}\\
\color{purple}{x = 1}\\
\end{equation}$
