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 527516: 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(x+5)=x^2+5x
(x-2)(2x+10)=x^2+5x+52
but I dont think Im setting it up right...
Can you help me set it up? And/or possibly give me the solution so I can figure out how to get to it for future problems?
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you're good so far...

expanding the 2nd eqn ___ 2x^2 + 6x - 20 = x^2 + 5x + 52 ___ x^2 + x - 72 = 0

factoring ___ (x + 9)(x - 8) = 0

x = 8
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