Why am I getting two different answers for this diff. equation? Completing the square vs quad form. With this differential equation, after seperating and integrating, and using the initial condition to solve for C, and then substituting that value of C into the general solution, I must solve for Y to get the particular solution.  Why am I getting two different solutions, where in method 1 I complete the square, and method 2, use the quadratic formula?
The ODE is:

$\frac{dy}{dx} = \frac{2x}{1+2y}$
The initial condition is:

$y(2) = 0$
Separating and integrating, I get the general soln.:

$y + y^2 = x^2 + c$
Plugging in the initial condition:

$0 + 0^2 = (2)^2 + c$
$c = -4$
Now, plugging in (-4) into the general solution:

$y + y^2 = x^2 - 4$
Now, method 1, completing the square:

$y^2 + y = x^2 - 4$
$(y + \frac{1}{2})^2 = x^2 - 4 + \frac{1}{4}$
$y = -\frac{1}{2} ^+_- (4x^2 - 15)^{\frac{1}{2}}$
Now, method 2, quad. formula:

$y = \frac{-1 ^+_- (1^2 - 4(1)(-x^2+4))^{\frac{1}{2}}}{2}$
$y = -\frac{1}{2} ^+_- \frac{1}{2}(4x^2-15)^{\frac{1}{2}}$
So, what did I do wrong with completing the square?  
The quad. formula is the same answer from the back of the
book.  There is a 1/2 in front of the square root, whereas with 
completing the square, the 1/2 is not there.   What did I do wrong?


$$\left( y+ \frac{1}{2}\right)^2= x^2-\frac{15}{4} \Rightarrow y+ \frac{1}{2}= \pm \sqrt{\left( x^2-\frac{15}{4}\right) }\Rightarrow y=-\frac{1}{2} \pm \sqrt{\left( x^2-\frac{15}{4}\right)} \\=-\frac{1}{2} \pm \sqrt{ \frac{1}{4}\left( 4x^2-15\right)}= -\frac{1}{2} \pm \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$$
EDIT: But since only $y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$ satisfies the initial condition $y(2)=0$, we reject $y(x)= -\frac{1}{2} - \frac{1}{2} \sqrt{\left( 4x^2-15 \right)}$ and so our only solution is:
$$y(x)=-\frac{1}{2}+ \frac{1}{2} \sqrt{4x^2-15}$$
