using $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
With:
a = 1, b = (-4) & c = (-3)
What do we do when we need to square root the -4 ?
using $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$
First of all, if you evaluate $\displaystyle x=b^2-4ac$, with the numbers you provided, you end up with $\displaystyle \sqrt{16-(-12)} = \sqrt{28} $ . So no problem here, i guess.
In the other case of a square root of a negative number, you will have to work with Complex numbers. Let us suppose you have to solve $\displaystyle x=\frac{-(-4)\pm\sqrt{-4}}{2}$. The next step is to write $\displaystyle \sqrt{-4}$ as $\displaystyle \sqrt{4i^2}$. If solve this root, you get $\displaystyle 2i$. Now we have $\displaystyle x=\frac{-(-4)\pm2i}{2}$
. Let's simplify all that and we get $\displaystyle x=2\pm i$
Greets,
Leslon