SOLUTION: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared

Question 527520: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared?
I tried...
(x-2)(2x+10)=x^2+5x+52
but I don't thin Im setting it up right...
Can you help me set it up?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
You are properly setting it up.
The original rectangle has a width that you called
and we should add a requirement that it should be a real number, and that

The length of the original rectangle is
Its area is AMP Parsing Error of [(x(x+5)=x^2+5x]: Invalid expression. Closing bracket expected =x^2+5\x at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187. .
The new rectangle has a length of
a width of
(so we are also going to ask for )
The area of the new rectangle is

Your equation transforms into
and into
(((x^2+x-72=0}}}
which can be solved by factoring, or by completing the square, or by using the quadratic formula.
One of the solutions for x is negative. The other solution is the width of the original rectangle, which you would need to use to find the area of the original rectangle.
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