I encounter that quadratic inequality can have two ways of graphing, line(s) with the area of solutions and parts of a parabola. How is that happen?

This is similar to your previous question.

If you are looking for $x$ such that 
\[x^2-4x\geq 14,\]
then 
\[x^2-4x- 14\geq 0.\]
Finding the roots using the quadratic formula, you get
\[x_1,x_2=\frac{+4\pm\sqrt{14+56}}{2}=\frac{+4\pm\sqrt{70}}{2}.\]
Hence the inequality is satisfied if 
\[ x\geq \frac{+4+\sqrt{70}}{2}   \text{or}    x\leq \frac{+4-\sqrt{70}}{2},\]
which are two line segments if graphed on $x$-axis. 

However, if you view this problem as the graph of the function $y=x^2-4x- 14$, then you want the portion of the graph that lies above $x$-axis, i.e. $y=x^2-4x- 14\geq 0.$

So it is the same problem, viewed from two different points of view. The first view is one dimensional where we only have the variable $x$, while in the second case we view the problem in two dimensions and have two variables $x,y$.