Question 548235
Let {{{x}}} be the width in inches.
The length, in inches, is {{{2x+2}}}.
The area, in square inches, is
{{{(2x+2)x=120}}} ---> {{{2x^2+2x=120}}} ---> {{{x^2+x=60}}} ---> {{{x^2+x-60=0}}}
You could have simplified the equation in a different way, using different steps, but the end result should be equivalent, and I believe my last equation is as nice and simple as you can get.
There are many ways to solve a quadratic equation like that one. You could try to factor the polynomial, but factoring does not work in this case (It only works when the solutions are nice rational numbers).
You could also use the quadratic formula, or try to "complete the square."
(To complete the square, I would start from {{{x^2+x=60}}}).
For this particular equation I did both, to make sure I could get the same answer two different ways. Both ways, I got
{{{x=(-1+sqrt(241))/2}}} as the positive answer. (We discard the negative answer, because lengths are always measured as positive numbers).
Then, the length of the rectangle is
{{{2x+2=2*((-1+sqrt(241))/2)+2=(-1+sqrt(241))+2=1+sqrt(241)}}} or about 16.5 inches.