Algebra I Help: Solving Quadratic Equations with Complex Number Solutions I

In this video, we explain how to solve a quadratic equation when the discriminant
(b^2-4ac) is negative. This means we end up with a negative square root, and no real solution. Once this occurs, we turn to our complex number system for an answer.

[Greenemath]

You can expand this to four terms and use "factoring by grouping". Let's begin:

z^2 + z(-2+i) - 2i = z^2 -2z + iz -2i = z(z-2) + i(z-2)
at this point we can factor out the (z-2):
(z+i)(z-2)
(z+i) = 0 if z = -i (z-2) = 0 if z = 2
That's where those two solutions come from. I hope this helps :)

[rekivan]

2:11 mistake !! the square never equals negative number my friend