**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 $x$.