Example  1.  Suppose $5x^2 - 20 = 0$. Find $x$.
Solution: Since $A \ne 0$, the equation can be solved so that $x =  \pm \sqrt{ - \frac C A}$  where $A=5$, and $C = -20$. So the solution is $ x =  \pm \sqrt{\frac {20} 5} =  \pm \sqrt{4} = \pm 2$.

Comment: We can consider the expression on the left hand side of the equation as a function of $x$ giving $  f(x) = 5x^2 - 20$ . Now the problem can be restated: to find a $x$ where $f(x) = 0$. This problem and its solution can be visualized both on the graph and the mapping diagram for the function $f$. 

You can change the value of $x$ by moving the point on the domain axis of the mapping diagram.
For the graph of $f$
: Find $y=0$ on the Y axis , then find the point(s) on the graph of $f$ with second coordinate $0$, determine it's first coordinates, $2$ and $-2$, and those are the desired values for $x$. To do this, look for the point(s) on the $X$ axis, the line $y=0$, where the axis intersects the graph of $f$
For the mapping diagram of $f$: Find $y=0$ on the target axis, then find point(s) $x$ on the source axis with the function arrow pointing to $0$.To do this, find the extreme point of $f$ on the mapping diagram, $x = -\frac B {2A} = 0$. Move symmetrically above and below that value by $\pm 2$, which are the desired values for $x$.