Question 635320
Let {{{x}}} be the width of the rectangle, in inches. 
The length of the rectangle( in inches) is {{{x+7}}}.
The area is the product of length times width, so it is
{{{x(x+7)}}} square inches.
The modified rectangle has a length (in inches) of {{{2(x+7)}}} ,
and a width (in inches) of {{{x-3}}} .
The area (in square inches) of the modified rectangle is
{{{(2(x+7))(x-3)}}} ,
and that new area is 90 square inches more than the area of the original rectangle.
So the equation, in its most complicated form (just in case the teacher wants that), is
{{{(2(x+7))(x-3)=x(x+7)+90}}}
Let's un-complicate and solve.
{{{(2(x+7))(x-3)=x(x+7)+90}}} --> {{{2((x+7)(x-3))=x(x+7)+90}}} (associative property)
(We are so familiar with he associative property that we probably could write it as
{{{2(x+7)(x-3)=x(x+7)+90}}} in the un-complicated no-parentheses form from the start).
Form there, we multiply the expressions with {{{x}}} to get
{{{2(x^2-3x+7x-21)=x^2+7x+90}}}
Collecting like terms, we get
{{{2(x^2+4x-21)=x^2+7x+90}}} 
We multiply further to get
{{{2x^2+8x-42=x^2+7x+90}}}
Then we add {{{-x^2-7x-90}}} to both sides (or subtract {{{x^2+7x+90}}}, same thing), to get
{{{2x^2+8x-42-x^2-7x-90=x^2+7x+90-x^2-7x-90}}}
That simplifies to our quadratic equation:
{{{x^2+x-132=0}}}
If we are good at factoring, we can easily figure that
{{{x^2+x-132=(x+12)(x-11)}}} and then we would re-write the equation as
{{{(x+12)(x-11)=0}}} , with solutions {{{x=-12}}} and {{{highlight(x=11)}}}
The width of the rectangle is {{{highlight(11)}}} inches and the length is {{{11+7=highlight(18)}}} inches.
 
Verification:
The area of the rectangle is (11 inches)(18 inches) = 198 square inches.
The length of the modified rectangle is
2(18 inches) = 36 inches, and the new width is
11 inches - 3 inches = 8 inches.
Then, the area of the modified rectangle is
(8 inches)(36 inches) = 288 square inches,
which is 90 square inches more than the original 198 square inches,
because 198+90=288.