<s>[INST] 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 $$\frac{dy}{dt} = \frac{t}{y-2}$$.
 [/INST] Complete the square in $y$:$$y^2-4y-t^2=C\\y^2-4y+4=t^2+C+4\\(y-2)^2=t^2+C\\y=2\pm\sqrt{t^2+C}$$
Note we can absorb the $4$ into our arbitrary constant $C$ which parameterizes our family of solutions.
From the original equation, notice: $$(y-2)\frac{dy}{dt}=t$$
Let $z=y-2$ so $dz=dy$ giving $$z\frac{dz}{dt}=t\\2z\frac{dz}{dt}=2t\\\int 2z\frac{dz}{dt}\,dt=\int 2t\, dt\\z^2=t^2+C\\z=\pm\sqrt{t^2+C}\\y-2=\pm\sqrt{t^2+C}\\y=2\pm\sqrt{t^2+C}$$
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