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I have a question in my excersise book:

By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$

My approach was to complete the square and rearrange to make $x$ the subject.

The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.

However I get a negative square root:

$$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
$$(sqrt {3}x - 2sqrt{3})^2 = -2$$
$$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
$$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
$$x = (2sqrt{3} +- sqrt {-2})/3$$

Bad formatting: $+-$ means either $+$ or $-$

Where have I gone wrong?

You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
$$
(sqrt 3x - 2sqrt3)^2 + 2
$$
then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.

You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.

Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.

In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.