(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 \(\boxed{1+i, 1-i}.\)

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