Answer
The solution set of this problem is $$\Big\{3\pm i\sqrt5\Big\}$$
Work Step by Step
$$x^2-6x+14=0$$
The equation is already in standard form, so the quadratic formula can immediately be used.
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
As $a=1, b=-6, c=14$
$$x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times14}}{2\times1}$$
$$x=\frac{6\pm\sqrt{36-56}}{2}$$
$$x=\frac{6\pm\sqrt{-20}}{2}$$
Now we rewrite $\sqrt{-20}=i\sqrt20=2i\sqrt5$
$$x=\frac{6\pm 2i\sqrt5}{2}$$
Then we simplify
$$x=3\pm i\sqrt5$$
The solution set of this problem is $$\Big\{3\pm i\sqrt5\Big\}$$
