Hallo Sarah,
2000·0.50.01·x = 100·20.01·x + 100
⇔ 2000·2-0.01·x = 100·20.01·x + 100
⇔ 2000·(20.01·x-1 = 100·20.01·x
Substitution z = 20.01·x
2000 * z-1 = 100 z + 100 | * z
2000 = 100 z2 + 100 z | - 2000 | : 100 | ↔
z2 + z - 20 = 0
pq-Formel → z = 4 oder z = - 5
20.01·x = 4 [ oder 20.01·x = - 5 < 0 entfällt ]
20.01·x = 22
0.01·x = 2 | * 100
x = 200
D(200) = S(200) = 500 (!)
Gruß Wolfgang