Answer:
[tex]y^{2}-5y=750[/tex]
[tex]750-y(y-5)=0[/tex]
[tex](y + 25)(y -30) = 0[/tex]
Step-by-step explanation:
Givens
Let's call [tex]w[/tex] the width and [tex]l[/tex] the length. According to the problem they are related as follows
[tex]w=l-5[/tex], because the width is 5 feet less than the lenght.
We know that the area of the room is defined as
[tex]A=w\times l[/tex]
Where [tex]A=750 ft^{2}[/tex]
Replacing the given area and the expression, we have
[tex]750=(l-5)l\\750=l^{2}-5l \\l^{2}-5l-750=0[/tex]
We need to find two number which product is 750 and which difference is 5, those numbers are 30 and 25.
[tex]l^{2}-5l-750=(l-30)(l+25)[/tex]
Using the zero property, we have
[tex]l=30\\l=-25[/tex]
Where only the positive number makes sense to the problem because a negative length doesn't make any sense.
Therefore, the length of the room is 30 feet.
Also, the right answers are the second choice where [tex]y=l[/tex], the third choice and the last choice.
