Question 5071
Starting with the quadratic formula:
 {{{x=(-b+-sqrt(b^2-4ac))/(2a)}}} and given that:
 {{{x=2+-sqrt(-1)}}} we can conclude that:
-b/2a = 2  Assuming that a is a non-zero integer, it must be 1, therefore:
-b/2 = 2
So, a = 1, and b = -4

Now we need to find c.  Looking at the radical contents, we can conclude that it is equal to -4.
Why? because {{{sqrt(b^2 - 4ac)/2}}} = -1  Substituting the values of a and b, we find:
{{{sqrt((-4)^2 - (4)(1)c)/2))}}} = {{{sqrt(-1)}}}
{{{sqrt((16-4c)/2)}}} = {{{sqrt(-1)}}}  Multiply both sides by 2. 
{{{sqrt(16 - 4c)}}} = 2{{{sqrt(-1)}}} Square both sides.
16 - 4c = 4(-1) Subtract 16 from both sides.
-4c = -20 Divide both sides by -4
c = 5

Now we can write the quadratic equation whose roots (zeros) are (2 +/- i)

x^2 - 4x + 5 = 0