Ineed um i dont need to to my work or check my answer just
as you can see, i can figure out the problems fine, its explaining how the solution is related to the situation. if you could , or if you see any problems with my work, it would be very appreciated. you for your time.

3.
area=legngth tiimes width
legnth is 10 more than 2 times width
l=10+2w
lw=area=(10+2w)w=10w+2w^2
area=3600
3600=10w+2w^2
so divide by 2
1800=5w+w^2
subtract 1800 from both sides
(note they used x instead of w)
0=w^2+5w-1800
then the facotred form is (w-40)(w+45)=0

0 product property measn if you have xy, then x and or y =0

therefor
w-40=0 and w+45=0
w=40 or -45
discard negative because negative legnth is not possible

explian how solution relates to situation
it is math so it is correct because we represented it corrrectly

the other solution is because of math and eliminate negative legnth becasue tha tis not possible
basically because we represented it correctly
not a very clear quetion to answer

basically 'because it represents the solution so we solved it and it is correct'

#3:
In the first paragraph, the equation is      x² + 5x - 1800 = 0 .
Here's where this equation comes from:
Call the width of the park 'x'.
Then, according to the question, the length is  (2x + 10).
When you multiply (width) x (length) to get the area, and
simplify the product, it turns out that   x² + 5x - 1800 = 0 .
But don't forget . . . 'x' started out being the width, and when
you figure out what 'x' is, that'll be the width of the dog park.

3.a). is factoring the equation, getting ready to solve it. 

3.b). is finding the solutions for 'x' from the factored form.
         Remember that every quadratic equation has two solutions.
         In this one,  they are  x = 40  and  x = -45 .

         'x' is the width of the dog park.  Can it be both values ?
         Can it be only one of them ?  Can it be 40 ?  How about -45 ?

3.c).  'x' is the width of the real dog park.  In the real world, there's
         no way the city engineers can build it with a negative width.
         That would be totally silly.  So the real width 'x'  =  40 yards.

         Now look way back at the question:
           "The length of the park is 10 more than twice the width."

         The width is  40 yards.
         Twice the width is  80 yards.
         10 more than that is  90 yards.
    ==>  The length is 90 yards.  

         Now we have (length = 90 yards) and  (width = 40 yards).

         How do we know if these are correct ?
         Well, let's use them to calculate the area.

                Area = (length) x (width)
                     = (90 yards) x (40 yards) = 3,600 square yards.

                 That's what the question said !
                 We got the right numbers !                  yay !


#4
This one is very similar to the dog park in many ways.

Again, in the first paragraph, the signs are missing from the
equation.  The correct equation is:          x² + 6x - 55 = 0 .  

This is also a tricky question, because some of the information
in the first paragraph is only there to confuse you.   That business                   about "... increase both the height and the width ... by 3 inches."                     
You don't need that !  It may come up in #5, I don't know.  But here
in #4, it never asks anything about it, and you don't need to know it.
All you need to know is that it's a "square image".  That does that
tell you ?  It tells you that the height and width are equal !
That's why the equation  x² + 6x - 55 = 0  "can be used to find the
height and width of the original image" .  They're both the same.

4.a).  is factoring the equation, getting ready to solve it.

4.b).  is finding the solutions for 'x' from the factored form.
         Remember that every quadratic equation has two solutions.
         In this one,  they are  x = 5  and  x = -11 .

         'x' is a dimension of the image.  Can it be either value ?
         Can it be only one of them ?  Can it be 5 ?  How about -11 ?

4.c).  'x' is a dimension of the real image.  In the real world,
         there's no way the graphic designer can start with an image
         that has a negative dimension.  That would be totally silly. 

        So the real dimension of the image is  5 inches.
        Since it's square, the length and width are both  5 inches.


Now I'll just bet that there's another problem coming along to
ask you what the area of the new, bigger image is going to be,
after he increases both the height and the width of the image
by 3 inches.         You don't need to worry about crossing that
bridge until you come to it.

Add up all the sides