Hi! Please help with my homework

(B): We subtract \(36\) from both sides, leaving us with \(\sqrt{-36}=6i, -\sqrt{-36}=-6i\) . Thus, the two solutions are \(\boxed{x=6i,\:x=-6i}.\)

(C): There are a few ways to solve this question:

Quadratic Formula:

Use \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)(thanks web2.0calc), and plugging in the values in the form \(ax^2+bx+c\), we attain \(x=\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}=1+i\) as one root, and \(x=\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}=1-i\) . Thus, the two roots are \(\boxed{1+i, 1-i}.\)

Completing The Square:

Our main goal is to get it into the form \(x^2+2ax+a^2=\left(x+a\right)^2\), so we have \(x^2-2x+\left(-1\right)^2=-2+\left(-1\right)^2\), and simplifying, we get \(\left(x-1\right)^2=-1=x=i+1\) and \(i-1\) . Thus, the two roots are \(\boxed{1+i, 1-i}.\)

Tertre,

Your LaTex IS nice.  And it is great that you are learning and practicing your skills here.

PLUS

Your answer is nice, I agree, but.....

This person has clearly stated that they want someone to do their homework for them.

Why are you facilitating such blatent cheating?  

It is not like this user can beat you up and take your lunch money if you do not help.

If that were the case I would understand better. 

This is harsh, you are a great member and I don't want you to feel bad.

I'm sure I have at times been guilty of the same 'sin'.

I just don't want this to be known primarily as a free cheat site. 

I want people to learn here, like you and I do.

Perhaps you could have part done it so that the user would at least have to finish the questions on their own.

JUMBO QUESTION

Using a circle in the complex plane  centered at (0,0) with a radius of 2 , the 6 roots of 64 are

2, 1 + √3 i , -1 + √3 i , -2, -1 - √3 i , 1 - √3 i 

Here's a pic of the solutions