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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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