Let's call the width of the rectangle "w".
According to the problem, the length of the rectangle is 12 meters more than 5 times the width, so we can write that as:
L = 5w + 12
And the area is given as 32 square meters, so we can write that as:
A = L * w = 32
Now we have two equations:
5w + 12 = L
L * w = 32
We can use the first equation to substitute the value of L in the second equation:
L = 5w + 12
32 = (5w + 12) * w = 5w^2 + 12w
Expanding the right side of the equation, we get:
32 = 5w^2 + 12w
Now we have a quadratic equation that we can solve for w. To do this, we can use the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / 2a
where a = 5, b = 12, and c = -32. Plugging in these values, we get:
w = (-12 ± √(12^2 - 4 * 5 * -32)) / (2 * 5)
w = (-12 ± √(144 + 640)) / 10
w = (-12 ± √(784)) / 10
w = (-12 ± 28) / 10
w = (16 / 10) or ( -40 / 10)
w = 1.6 or -4
Since the width cannot be negative, the width must be 1.6 meters.