diff --git a/contents/solutions/AK105.md b/contents/solutions/AK105.md
index a02ff01..37bae8b 100644
--- a/contents/solutions/AK105.md
+++ b/contents/solutions/AK105.md
@@ -16,7 +16,7 @@ $d = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E}$
 
 # Einsetzen
 
-$d_{{40}\degree} = \frac{\sqrt{30 \Omega \cdot P_{{40}\degree}}}{E} = \frac{\sqrt{30 \Omega \cdot 0,25 \cdot P_{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{30 \Omega \cdot P_{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{30 \Omega \cdot P_{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{30 \Omega \cdot P_{EIRP}}}{E} = 0,5 \cdot d$
+$d_{{40}\degree} = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{{40}\degree}}}{E} = \frac{\sqrt{\qty{30}{\ohm} \cdot 0,25 \cdot P_{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_{EIRP}}}{E} = 0,5 \cdot d$
 
 # Interpretation der Formel