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?
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 $$x = 2C \left(\sqrt{\left(1 + \dfrac{y}{2C} \right)} - 1\right)$$
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
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 $$ x_2 = -2C - \sqrt{2C(2C+y)} $$
