Some weird equations In our theoreticall class professor stated that from this equation $(C = constant)$
$$
x^2 + 4Cx - 2Cy = 0
$$
we can first get:
$$
x = \frac{-4C + \sqrt{16 C^2 - 4(-2Cy)}}{2}
$$
and than this one:
$$
x = 2C \left[\sqrt{1 + \frac{y}{2C}} -1\right] 
$$
How is this even possible?
 A: Factor out the $16 C^2$ from the square root and simplify.
A: This comes from the quadratic formula. See here for more details.
In your case, $$x^2 + 4Cx -2Cy = x^2 + 4Cx +(2C)^2 - (2C)^2-2Cy = (x+2C)^2 - (2C)^2 \left(1 + \dfrac{y}{2C} \right) = 0$$
Hence, we have that
$$(x+2C)^2 = (2C)^2 \left(1 + \dfrac{y}{2C} \right) \implies x + 2C = \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)}$$
Hence, $$x = -2C \pm 2C \sqrt{\left(1 + \dfrac{y}{2C} \right)} = 2C \left( \pm \sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$
If you have a constraint that $x$ has the same sign as $C$ (for instance if $x,C > 0$), then $$x = 2C \left(\sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$
A: The secon one is just applying the general formula for second grade equations for x. From the second to the third you have to get out of the square root a common factor of $4C^2$, that becomes $2C$ when coming out of the root
A: Here's the algebra:
$$x^2 + 4Cx - 2Cy = (x+2C)^2-4C^2 - 2Cy = 0 $$
Thus:
$$
(x+2C)^2 = 4C^2 + 2Cy = 2C(2C+y).
$$
Take square roots:
$$
x_1 = -2C + \sqrt{2C(2C+y)} =\frac{-4C + \sqrt{16C^2 +8Cy}}{2}$$
and
$$
x_2 = -2C - \sqrt{2C(2C+y)}
$$
