1990 European Under-18 Football Championship

The 7th U-18 European Football Championship was held in Hungary from July 22 to 29, 1990 . The winners were again the Soviet Union with a 4-2 victory on penalties against Portugal . None of the German-speaking teams ( Federal Republic of Germany , GDR , Austria , Switzerland ) could qualify.

mode

The eight qualified teams played for the title in a knockout system . The quarter-final winners reach the semi-finals, the semi-final winners the final. The semi-final losers play for third place. The quarter-final losers play for two free places for the 1991 World Junior Cup . The semi-finalists also qualify for the World Cup.

Attendees

The following teams took part in the tournament:

Belgium  England  Ireland  Portugal

Sweden  Soviet Union (defending champion)  Spain  Hungary (host)

Venues

The games were played in the cities of Békéscsaba , Debrecen , Gyula , Nyíregyháza and Szarvas .

The tournament

Quarter finals

July 24, 1990 in Nyíregyhaza
England	-	Belgium	1: 1 n.V., 5: 4 i. E.
July 24, 1990 in Debrecen
Soviet Union	-	Sweden	3-0
July 24, 1990 in Békéscsaba
Portugal	-	Hungary	1: 1 n.V., 3: 1 i. E.
July 24, 1990 in Gyula
Spain	-	Ireland	3-0

World Cup qualification

July 26, 1990 in Nyíregyháza
Sweden	-	Belgium	6-0
July 26, 1990 in Szarvas
Ireland	-	Hungary	1-0

Semifinals

July 26, 1990 in Gyula
Portugal	-	Spain	2: 1
July 26, 1990 in Debrecen
Soviet Union	-	England	3: 1

Game for third place

July 29, 1990 in Békéscsaba
Spain	-	England	6-0

final

July 29, 1990 in Békéscsaba
Soviet Union	-	Portugal	0: 0 a.d., 4: 2 i. E.

decisions

The Soviet Union became European U-18 champions for the second time.

The winning eleven: Pomazun - Krbasjan, Busmanow, Minko, Mamcsur - Guscsin (Pohlabejew), Grisin, Csaran - Mandreko (Kandurow), Lukin, Scherbakow

In addition to the Soviet Union, England, Ireland, Portugal, Sweden and Spain qualified for the 1991 World Youth Championship .

Web links

• U-18 European Championship 1990 on rsssf.com (engl.)