Are you familiar with how to complete the square? If so, then we rearrange your original equation,
$\displaystyle 3(x^2 - 4\sqrt{3}) - 8x(\sqrt{3} - 1) + 19 = 0$
into:
$\displaystyle 3x^2 - 8(\sqrt{3} - 1)x + 19 - 12\sqrt{3} = 0$
Factor out the 3:
$\displaystyle 3\left[x^2 - \frac{8(\sqrt{3} - 1)x}{3}\right] + 19 - 12\sqrt{3} = 0$
And complete the square:
$\displaystyle 3\left[\left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2\right] + 19 - 12\sqrt{3} = 0$
Then some algebraic manipulation gets you (incoming wall-of-step-by-step-text):
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 - \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 = \frac{12\sqrt{3} - 19}{3}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \left (\frac{4(\sqrt{3} - 1)}{3}\right )^2 + \frac{12\sqrt{3} - 19}{3}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{16(4-2\sqrt{3})}{9} + \frac{36\sqrt{3} - 57}{9}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{64 - 32\sqrt{3} + 36\sqrt{3} - 57}{9}$
$\displaystyle \left (x-\frac{4(\sqrt{3}-1)}{3}\right )^2 = \frac{7 + 4\sqrt{3}}{9}$
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\sqrt{\frac{7 + 4\sqrt{3}}{9}}$
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{7 + 4\sqrt{3}}}{3}}$
From here, notice that:
$\displaystyle 7+4\sqrt{3} = (2+\sqrt{3})^2$
Substituting that into the equation, we get:
$\displaystyle x-\frac{4(\sqrt{3}-1)}{3} = \pm\frac{\sqrt{(2+\sqrt{3})^2}}{3}}$
So,
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}\pm\frac{2+\sqrt{3}}{3}}$
Now, evaluate both solutions for x:
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}+\frac{2+\sqrt{3}}{3}}=\frac{5\sqrt{3}-2}{3}$
$\displaystyle x=\frac{4(\sqrt{3}-1)}{3}-\frac{2+\sqrt{3}}{3}}=\frac{3\sqrt{3}-6}{3}=\sqrt{3}-2$
Voila! Done!
If you're not sure how to complete the square, I'm sure there are other methods to doing this, such as using the quadratic equation as DavidM suggested. The "tricky" part is recognising that $\displaystyle 7+4\sqrt{3}$ can be expressed as a square in itself.