Solve a quadratic equation Given
$$x^2+\frac{2D}{k}x=\frac{2DC_{o}}{C_{l}}(t+\tau)$$
How do I get to
$$x=\frac{D}{k}\left[\sqrt{1+\frac{2C_{o}k^2(t+\tau)}{DC_{l}}}-1\right]$$
Thanks!
 A: If you have:
$$x^2+2ax=c$$
You could sum $a^2$ in both sides to obtain:
$$x^2+2ax+a^2=c+a^2$$
$$(x+a)^2=c+a^2$$
Then is the same:
$$x^2+\frac{2D}{k}x=\frac{2DC_{o}}{C_{l}}(t+\tau)$$
Sum $\frac{D^2}{k^2}$ in both sides:
$$x^2+\frac{2D}{k}x+\frac{D^2}{k^2}=\frac{2DC_{o}}{C_{l}}(t+\tau)+\frac{D^2}{k^2}$$
$$\left(x+\frac{D}{k}\right)^2=\frac{2DC_{o}}{C_{l}}(t+\tau)+\frac{D^2}{k^2}$$
Taking root:
$$\left(x+\frac{D}{k}\right)=\pm\sqrt{\frac{2DC_{o}}{C_{l}}(t+\tau)+\frac{D^2}{k^2}}$$
$$x=\pm\sqrt{\frac{2DC_{o}}{C_{l}}(t+\tau)+\frac{D^2}{k^2}}-\frac{D}{k}$$
Taking factor $\frac{D^2}{k^2}$ in the square root:
$$x=\pm\sqrt{\frac{2D^2C_{o}k^2}{DC_{l}k^2}(t+\tau)+\frac{D^2}{k^2}}-\frac{D}{k}$$
$$x=\pm\left|\frac{D}{k}\right|\sqrt{\frac{2C_{o}k^2}{DC_{l}}(t+\tau)+1}-\frac{D}{k}$$
And if you assume $\frac{D}{k}>0$ 
$$x=\frac{D}{k}\left[\pm\sqrt{\frac{2C_{o}k^2}{DC_{l}}(t+\tau)+1}-1\right]$$
The positive ($+$)solution is what you want (it seems like you are solving a physics problem that could explain the choise of the positive solution)
