If you want to multiply (2*x + 5) and (2*x - 5), you can do this using the following steps:
(2*x + 5) * (2*x - 5) = 4*x^2 - 10*x + 10*x - 25 = 4*x^2 - 25
The correct result is 4*x^2 - 25.
Example:

This suggests two solutions,

and

.
However, upon plugging these solutions back into the equation, you get

which checks out, but

does not because

is defined only for

(assuming you're looking for real solutions only). So, we call

an extraneous solution, and the complete solution set (over the real numbers) is

.
Step-by-step explanation:
1) A number <em>m</em> is at least four
<em>m ≥ 4 ['at least' = ≥ symbol]</em>
<em />
2) 4x ≤ 20x; x=2
4(2) ≤ 20(2)
8 ≤ 40
[Im not sure if this is a solution or and inequality :/ ]
3) x < 5
--> On the line, draw the number 0, and the number five. 'x < 5' also means 'x is less that five.' So, draw a dot over the line before 5, and make an arrow going to the left of the page,
<-----------------------------------o 5
<---|---|---|---|---|---|---|---|---|---|---|---|---|---|--->
0
Hopefully this helps! :)
The initial value is how much he starts out with, and it says he has $32
So the initial value is B, $32