we have
[tex]2x^{2} +x-6[/tex]
we know that
The area of a rectangle is equal to
[tex]A=L*W[/tex]
where
L is the length side of the rectangle
W is the width side of the rectangle
In this problem
Equate the area to zero and Find the roots of the quadratic equation
[tex]2x^{2} +x-6=0[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]2x^{2} +x=6[/tex]
Factor the leading coefficient
[tex]2(x^{2} +0.5x)=6[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]2(x^{2} +0.5x+0.25^{2})=6+0.125[/tex]
[tex]2(x^{2} +0.5x+0.0625=6.125[/tex]
Rewrite as perfect squares
[tex]2(x+0.25)^{2}=6.125[/tex]
[tex](x+0.25)^{2}=3.0625[/tex]
Square root both sides
[tex](x+0.25)=(+/-)\sqrt{3.0625}[/tex]
[tex](x+0.25)=(+/-)1.75[/tex]
[tex]x1=-0.25+1.75=1.50[/tex]
[tex]x2=-0.25-1.75=-2[/tex]
so
[tex]2x^{2} +x-6=2(x+2)(x-1.50)[/tex]
[tex]2(x+2)(x-1.50)=(x+2)(2x-3)[/tex]
[tex]length=(2x-3)\ feet[/tex]
[tex]width=(x+2)\ feet[/tex]
therefore
the answer is the option
[tex]length=(2x-3)\ feet[/tex] ; [tex]width=(x+2)\ feet[/tex]
