Question 200644
I don't have a diagram of this situation but we can muddle through without one.
First, the big rectangle has an area of:
{{{A[1] = (2x-2)(3x)}}}
{{{A[1] = 6x^2-6x}}} now we'll subtract the area of the three ({{{3(x*(x-1))}}}) rectangular holes whose total area is: 
{{{A[2] = 3x^2-3x}}} to get:
{{{A[1]-A[2] = 6x^2-6x-(3x^2-3x)}}}
{{{A[t] = 3x^2-3x}}} and this, the problem tells us is 90, so...
{{{3x^2-3x = 90}}} Subtract 90 from both sides.
{{{3x^2-3x-90 = 0}}} Factor out a 3 to facilitate the calculation.
{{{3(x^2-x-30) = 0}}} which means that...
{{{x^2-x-30 = 0}}} Factor this quadratic equation.
{{{(x+5)(x-6) = 0}}} so...
{{{highlight(x = -5)}}} or {{{highlight_green(x = 6)}}} Discard the negative solution as the length, x, can only be positive.