Something seems a bit off though.
The roots of the equation are $\displaystyle \frac{-6\pm\sqrt{6^2-4\cdot1\cdot13}}{2\cdot 1}$, which you wrote as -6+-√6^2-4(1)(13)/2(1). As you know, in the standard order of operations division comes before subtraction, so your formula is $\displaystyle -6\pm\sqrt{6^2-\frac{4\cdot1\cdot13}{2\cdot1}$. (In the text formula, it is also not clear where the square root ends.) And what about -6+-√16/2, which means $\displaystyle -6\pm\sqrt{\frac{16}{2}}$ or $\displaystyle -6\pm\frac{\sqrt{16}}{2}$, but not $\displaystyle \frac{-6\pm\sqrt{16}}{2}$?
In any case, the main mistake is that the discriminant is negative.