If the length of a rectangle is 11 yds more than twice the width and the area of the rectangle is 63yds^2 what is the dimension of the rectangle?

Width = w
length = 2w + 11
area = length * width = (2w+11)*w = 63
w = 3 gives area = 3*17 = 51. w=4 gives area = 4 * 19 = 96. So w is somewhere between 3 and 4.
2w^2 + 11w - 63 = 0
That's a quadratic equation. We can test and find out if it's easily solved by factoring. Otherwise the quadratic formula can solve it.

(2w - a)(w + b) = 0
a*b = 63
2b - a = 11
(or, a - 2b = 11)

Let's factor 63.
63 = 3*21 = 3*3*7
1 and 63 (that doesn't work).
7 and 9: 14-9=5 nope, 18-7=11 yes. So b=9, a=7 will work.

(2w - a)(w + b) = 0 becomes
(2w - 7)(w + 9) = 0
That means w can be 7/2, or -9. -9 is obviously wrong, so let's check 7/2. The area of the rectangle was w*(2w+11) That's 3.5 * (7+11) = 3.5*18 = 63. So the rectangle's width is 3.5, and its length is 18.