Quadratic equation with several variables How does 
$$y^{2} - 4y -t^{2} - C = 0$$
Become 
$$y = 2 \pm \sqrt{t^{2} +2C + 4}$$
I know its the quadratic formula but I dont know how it got it that point
The original equation is $$\frac{dy}{dt} = \frac{t}{y-2}$$.

The equation 
$$\frac{dy}{dt} = \frac{t}{y-2}$$
can be seen in the form
\begin{align}
(y-2) \, dy = t \, dt 
\end{align}
and upon integration of both sides becomes
\begin{align}
\frac{y^{2}}{2} - 2 y = \frac{t^{2}}{2} + \frac{c_{0}}{2}
\end{align}
or $y^{2} - 4 y -(t^{2} + c_{0}) = 0$. Solving this quadratic equation the value of $y$ is of the form
\begin{align}
y(t) = 2 \pm \sqrt{t^{2} + 4 + c_{0}}.
\end{align}
