(16 M, 33 M) Let the breadth of the plot be $x \ m$. Then the length is $(2x + 1) \ m$.
Given that the area is $528 \ m^2$,we have $x(2x + 1) = 528$,which simplifies to $2x^2 + x - 528 = 0$.
This is a quadratic equation of the form $ax^2 + bx + c = 0$,where $a = 2, b = 1, c = -528$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-528)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 4224}}{4} = \frac{-1 \pm \sqrt{4225}}{4} = \frac{-1 \pm 65}{4}$.
This gives two possible values for $x$: $x = \frac{64}{4} = 16$ or $x = \frac{-66}{4} = -16.5$.
Since the breadth cannot be negative,we take $x = 16 \ m$.
Therefore,the breadth is $16 \ m$ and the length is $2(16) + 1 = 33 \ m$.