Dear Colleagues

I have the following question to find a solution to:
The length of a rectangle exceeds its breadth by 4 centimetres.  If the length were halved and the breadth 
increased by 5 centimetres, the area would be decreased by 35 square centimetres.  Find the length of the rectangle.

I do not know if I can draw in this website, but whilst trying to find the solutions, I drew 2 rectangles: 
the one on the left had the dimensions (x)(x+4) for width and length respectively, 
the one on the right had dimensions ((x+4)/2)(x+5) for length and width respectively.
The question then reads - for me - [((x+4)/2)(x+5)] - 35 = x(x+4)   <<<---===
Multiplying the left hand by 2 to clear the fraction, we get
[(x+4)(2x+10)]-35 = x(x+4)
and after multiplying out we get
[2x^2+10x+8x+40]-35 = x^2+4x
Collecting like terms gives us
2x^2+18x+5 = x^2+4x
which simplifies to
x^2+14x+5=0
We cannot factorise, so using the quadratic formula we get
a = 1, b= 14, c = 5
and putting all of this into the formula we get

x = (-b+/- sqrt of b^2-4ac)/2
which gives
x = (-14 +/- sqrt (14)^2-(4)(1)(5))/2

which gives us, finally, x = (-14 +/-13.23)/2

I am assuming that we need a real number for the length of the side of the rectangle, 
so do not wish to get into imaginary numbers...
Where have I gone wrong?

The first error you made was forming an incorrect equation. 
The easiest way to do this problem is to let the variable be what's asked for.
In other words, let the original length (requested value) be L
Then the breadth is L - 4
Halving the length gives the new length, 
Increasing the breadth by 5 gives the new breadth, L - 4 + 5, or L + 1
Original area: L(L - 4)
Given that the original area is reduced by 35, we get: 
                 ------ Multiplying by LCD, 2
           
       (L - 14)(L + 5) = 0
                L - 14 = 0            or      L + 5 = 0 
Original length, or       or      L = - 5 (ignore)