Explanation:

Let the length of the playground be x and breadth be y.
Given, area of the rectangular playground = 100 square metre
∴  x × y = 100       ...(1)
30 m barbed wire is used to fence 3 sides. If one length side is not fenced using wire then we can say,
x + 2y = 30       ...(2)
⇒  x + 2 × 100/x = 30
[From eq. (1)]
⇒  x² + 200 = 30x
⇒  x² – 30x + 200 = 0
⇒  x² – 20x – 10x + 200 = 0
⇒  x(x – 20) – 10(x – 20) = 0
⇒  (x – 20)(x – 10) = 0
⇒  x = 20 or 10
Now, From eq. (1),
y = 5 or 10
As the playground is rectangular,
x = y = 10 is not possible,
so x = 20 m and y = 5 m.
Intuitive Approach :
The only two possible combinations of two sides of a rectangle, the are of which is 100 square metre are is 20 m × 5 m and 25 m × 4 m.
Assuming 20 m × 5 m to be true, we can validate our choice by putting the values in the following given condition.
x + 2y = 30 m
⇒  20 + 2 × 5 = 30
⇒  30 = 30
L.H.S. = R.H.S.
Hence, it is evident that the pair of sides (20 m × 5 m) is correct.