The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle?

If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?
2x^2 + 3x - 119 = 0
3x^2 + 3x - 119 = 0
6x^2 - 119 = 0

Question

The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle?

If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?
2x^2 + 3x - 119 = 0
3x^2 + 3x - 119 = 0
6x^2 - 119 = 0

anonymous · Answer

$$Area=l*w$$|dw:1337705707236:dw|

anonymous · Answer

so it is the 1st one

anonymous · Answer

$$(2x+3)x=119$$$$2x^2+3x=119$$ $$2x^2+3x-119=0$$ yup thats right.

anonymous · Answer

ok thanks