We will use the quadratic formula to solve for the x values of this quadratic. [tex] x=\frac{3+/-\sqrt{3^2-4(2)(6)}}{2(2)} [/tex]. Simplifying that a bit gives us [tex] x=\frac{3+/-\sqrt{9-48}}{4} [/tex] and [tex] x=\frac{3+/-\sqrt{-39}}{4} [/tex]. We can't leave that negative under there. So we use the fact that -1 = i^2 and make the substitution. [tex] x=\frac{3+/-\sqrt{i^2*39}}{4} [/tex]. i^2 is a perfect square that we can pull out, leaving us with [tex] x=\frac{3+/-i\sqrt{39}}{4} [/tex]. The first of your choices above is the one that you want.
