\(\begin{array}{lccllll}
1x^2 &- 16x &+ 64& = 0\\
\uparrow &\uparrow &\uparrow \\
a&b&c
\end{array}\qquad discriminant\to b^2-4\cdot a\cdot c\)
if the discriminant is
negative <-- two complex solutions
positive <-- two real solutions
0 <-- one solution