The length of a rectangle is 59 inches greater than twice the width. if the diagonal is 2 inches more than the length, find the dimensions of the rectangle.

The question is asking us to find the dimensions of the rectangle, which would be the length and width. So, to find this, we must first state our givens, as it is Geometry.
Given: Length of rectangle = 59 + twice the width, diagonal = 2 inches longer than the width

Let's first translate all our givens to numbers. We'll start off by assigning variables that are easy to work with (x, y and z). 
x = width
y = length
z = diagonal
Now that we have done that, we need to translate all our givens into numbers. Here is how that would look like:
y = 2x + 59      ←59 plus twice the width (x)
z = y + 2           ←Diagonal = 2 inches more than width
If we draw a diagram, we can see that the diagonal, length, and width all create a right triangle, which means that we can use the Pythagorean Theorem. By using right triangle postulates and theorems, we can deduce that the diagonal is the hypotenuse. Here is what our setup looks like:
x² + y² = z²
Now, all we need to do is plug in the expressions we created for y and z:
x² + (2x + 59)² = [2 + (2x + 59)²]
When we solve for x, we get x = 20. Now, we just plug the x value back into the y equation to get 99. Therefore, the length equals 99 inches and the width equals 20 inches. Hope this helps and have a great day!

The answers are 20,30,80,180,960,890,1230,4560. on the first row