diff --git a/contents/solutions/AK105.md b/contents/solutions/AK105.md
index 73f5554..2d262c2 100644
--- a/contents/solutions/AK105.md
+++ b/contents/solutions/AK105.md
@@ -16,7 +16,7 @@ $d = \frac{\sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E}$
 
 ## Einsetzen
 
-$d_{{40}^\circ} = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{{40}^\circ}}}{E} = \frac{\sqrt{\qty{30}{\ohm} \cdot 0,25 \cdot P_\mathrm{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = 0,5 \cdot d$
+$d_{\qty{40}{^\circ}} = \frac{\sqrt{\qty{30}{\ohm} \cdot P_{\qty{40}{^\circ}}}}{E} = \frac{\sqrt{\qty{30}{\ohm} \cdot 0,25 \cdot P_\mathrm{EIRP}}}{E} = \frac{\sqrt{0,25} \cdot \sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = \sqrt{0,25} \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = 0,5 \cdot \frac{\sqrt{\qty{30}{\ohm} \cdot P_\mathrm{EIRP}}}{E} = 0,5 \cdot d$
 
 ## Interpretation der Formel