What is the general solution of $\tan^2x- \sqrt3 \tan x -\tan x= -\sqrt3$ I require some help finding the answer to the above equation. I have tried using the quadratic formula, however I couldn't get the answer $-\sqrt3$.

Ok, so you have
$$
\tan^2 x - \sqrt{3}\tan x - \tan x = -\sqrt{3},
$$
this is
$$
\tan^2 x - \left(\sqrt{3}+1\right)\tan x + \sqrt{3} = 0.
$$
Substituting $\tan x=y$, you get
$$
y^2 - \left(\sqrt{3}+1\right)y + \sqrt{3} = 0
$$
We get the following values for $y$:
$$
\begin{split}
y=\frac{\sqrt{3}+1\pm\sqrt{\left(-\sqrt{3}-1\right)^2-4\sqrt{3}}}{2} & = %
\frac{\sqrt{3}+1\pm\sqrt{4-2\sqrt{3}}}{2}=\\[1em]
& = \frac{\sqrt{3}+1\pm\left(\sqrt{3}-1\right)}{2}=
\left\{\begin{array}{ll}\sqrt{3}\\1\end{array}\right.
\end{split}
$$
So either $\tan x=\sqrt{3}$ or $\tan x=1$.
In the first case
$$
x=\arctan\left(\sqrt{3}\right)=\frac{\pi}{3}\quad\vee\quad x=\arctan\left(\sqrt{3}\right)+\pi=\frac{4}{3}\pi.
$$
In the second case instead
$$
x=\arctan(1)=\frac{\pi}{4}\quad\vee\quad x=\arctan(1)+\pi=\frac{5}{4}\pi.
$$
It follows that the set of solutions is
$$
x=\frac{\pi}{4}+k\pi\quad\vee\quad x=\frac{\pi}{3}+k\pi,
$$
for all $k\in\Bbb Z$.
