How does $x^2-8x+17=0$ have nonreal solutions? 
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.

Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17 
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?

Note that $\sqrt{-1}=i$$$x^2-8x+17=0 \implies x=\frac{8\pm \sqrt{64-68}}{2}=\frac{8+\sqrt{-4}}{2}=4\pm i$$
So (A) option is correct.
