Question 1059919
The zeros are 3 +/- 2i, since complex roots come as conjugate
This is the result of -b+/-sqrt(b^2-4ac) all divided by 2a
Therefore, b=-6, because it will be divided by 2 at the end
b^2-4ac=36-4*c, and that has to equal -16.  The sqrt (-16) is +/-4i, and dividing that by 2 will make 2i
Therefore, 4c=52, and c=13.
x^2-6x+13=0
{{{graph(300,300,-10,10,-10,10,x^2-6x+13)}}}