Introduction:
In this problem, we are given a concave mirror with a focal length of 20 cm and a magnification of 3 cm. We need to calculate two possible distances of the object from the mirror.

Given:
Focal length (f) = 20 cm
Magnification (m) = 3 cm

Formula:
The formula relating object distance (u), image distance (v), and focal length (f) is:
1/f = 1/v - 1/u

Approach:
To find the two possible distances of the object from the mirror, we can use the formula mentioned above. We will consider two cases: one where the image is real and the other where the image is virtual.

Case 1: Real Image
When the image is real, the magnification (m) is positive. Let's assume the object distance (u) as a positive value.

Step 1:
Given, magnification (m) = 3 cm
m = v/u
3 = v/u

Step 2:
Using the formula, 1/f = 1/v - 1/u, substitute the values of f and v:
1/20 = 1/v - 1/u

Step 3:
Substitute the value of v from Step 1 into the equation in Step 2:
1/20 = 1/(3u) - 1/u

Step 4:
Solve the equation to find the value of u:
1/20 = (u - 3u)/(3u^2)
1/20 = -2u/(3u^2)
3u^2 = -40u
3u^2 + 40u = 0
u(3u + 40) = 0

Step 5:
Solve the equation to find the two possible values of u:
u = 0 (not valid)
3u + 40 = 0
3u = -40
u = -40/3
u ≈ -13.33 cm

Therefore, the two possible distances of the object from the mirror when the image is real are 0 (not valid) and approximately -13.33 cm.

Case 2: Virtual Image
When the image is virtual, the magnification (m) is negative. Let's assume the object distance (u) as a negative value.

Step 1:
Given, magnification (m) = 3 cm
m = v/u
-3 = v/u

Step 2:
Using the formula, 1/f = 1/v - 1/u, substitute the values of f and v:
1/20 = 1/v - 1/u

Step 3:
Substitute the value of v from Step 1 into the equation in Step 2:
1/20 = 1/(-3u) - 1/u

Step 4:
Solve the equation to find the value of u:
1/20 = (-u + 3u)/(-3u^2)
1/20 = 2u/(-3u^2)
-3u