Complex quadratic equation always comes out as wrong

Your mistake is that the discriminant should be $12^2 - 4\times 4\times 19 = -160$, not $-260$, and then dividing top and bottom by $4$ gives the correct answer.

When the $x$ coefficient is even, say $ax^2+2bx+c=0,$ there is an equivalent version of the quadratic equation which automatically cancels that extra factor of $4.$

Instead of the usual discriminant $D=b^2-4ac$ with this setup one calculates what might be called the "other" discriminant (for want of a better word) $E=b^2-ac$ as the thing under the radical. As usual if that's negative the complex number $i$ is used when square root of discriminant is taken.

Then the roots are $$\frac{-b \pm \sqrt{E}}{a}.$$

For the re-written example you have: $4z^2-2(-6)z+19=0,$ we get $E=(-6)^2-4\cdot 19=36-4\cdot 19=36-76=-40.$ Then roots are $(6 \pm \sqrt{-40})/12.$ This simplifies again in this case.