# SOLUTION: The length of a rectangle exceeds twice its width by 3inches. If the area is 10 square inches, find the rectangle's dimensions. Round to the nearest tenth of an inch.

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 Click here to see ALL problems on Geometry Word Problems Question 79666: The length of a rectangle exceeds twice its width by 3inches. If the area is 10 square inches, find the rectangle's dimensions. Round to the nearest tenth of an inch.Answer by ptaylor(2048)   (Show Source): You can put this solution on YOUR website!Let x=width Then 2x+3=length Area of a rectangle =Length times Width or A=L*W So our equation to solve is: 10=x(2x+3) get rid of parens 10=2x^2+3x subtract 10 from both sides and rearrange 2x^2+3x-10=0 quadratic in standard form; solve using the quadratic formula and So ---------------------discount negative lengths and inches-----------------width inches----------- length CK A=10=(1.6)*(6.2) ~10=10 Hope this helps----ptaylor