Help solving for x using i (complex/imaginary numbers)

Solve for the real number x

x-4i = (2-i)^2

I missed this lesson in class and don't really understand what it means or how to solve it. All I know is i is imaginary or a complex number

The answer is x=3

x-4i=(2-i)(2-i)
x-4i=4-2i-2i+i^2
x-4i=4-4i+i^2
x=4+i^2
i^2= -1
x=4-1
x=3

Well, first off i=square root of -1, which is an imaginary number, also i²=-1. A complex number in standard form is a+bi, where a and b are both real numbers.
Here are the steps to solve the equation:
1. Use the distributive property on the right side of the equation. (2-i)² = (2-i)(2-i) = (4-4i+i²) = (4-4i-1) = 3-4i. So, x-4i=3-4i.
2. Add 4i to both sides of the equation: x-4i+4i=3-4i+4i = x=3


You can use the website http://www.classzone.com/cz/index.htm to find your textbook and go through the lesson online. You'll probably have to create an account on the website in order to do that.

first things first you must know some things
i= sqrt-1 theres nothing you can do about that
i^2= -1
i^3= -i or -sqrt-1
i^4= 1
ok? lets get started

x-4i = 4-2i-2i+i^2
x-4i = 4-4i-1 all thats left is to get x alone and combine like terms
x = 3

x - 4i = (2 - i)^2

=> x - 4i = 4 - 4i + i^2

=> x - 4i = 4 - 4i - 1

=> x - 4i = 3 - 4i

equate real and imaginary parts

x = 3

i = √(–1), or i² = –1

(2 – i)² = 4 – 4i – 1

x = 3