the dimensions of a rectangle can be expressed as x+6 and x-2. if the area of the rectangle is 65in^2, find the dimensions of the rectangle.

[tex]\bf \textit{area of a rectangle}\\\\ A=length\cdot width\qquad \boxed{A=65~in^2}\qquad 65=(\stackrel{length}{x+6})(\stackrel{width}{x-2}) \\\\\\ 65=x^2+4x-12\implies 0=x^2+4x-77 \\\\\\ 0=(x+7)(x-11)\implies x= \begin{cases} -7\\ \boxed{11} \end{cases}[/tex]

it cannot be -7, because that would give a negative value for either dimension, and a dimension for the rectangle cannot be negative.

so, the dimensions would be (11) + 6   and   (11) - 2.