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Romelu Lukaku scored four goals, his first after just 30 seconds, in Everton's 6-3 demolition of Bournemouth in the Premier League on Saturday.

By Associated Press
February 04, 2017

LIVERPOOL, England (AP) — Romelu Lukaku scored four goals, his first after just 30 seconds, in Everton's 6-3 demolition of Bournemouth in the Premier League on Saturday.

Almost immediately after the opening whistle, Lukaku breached Bournemouth's defense with the Belgium striker weaving past defenders to curl in a superb finish. It was the quickest goal scored by Everton at Goodison Park in the Premier League and also the joint fastest in the top flight this season. Chelsea's Pedro also scored after 30 seconds against Manchester United in October.

Lukaku then turned provider for Everton's second goal midway through the first half although it had a touch of luck. Lukaku did all the hard work, cutting the ball back from the right to James McCarthy whose shot struck Steven Cook before bouncing back off the Everton man's knee and into the net.

Having conceded twice just a quarter of the way through the game, Bournemouth could not afford to make any further mistakes at the back but Simon Francis then gifted Everton a third goal five minutes later.

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Francis gave away the ball to Lukaku who then only had Artur Boruc to beat. He did that comfortably, lifting the ball over the Polish 'keeper.

Bournemouth made a fight of it after halftime and threatened a comeback when Joshua King scored twice to make it 3-2 with 20 minutes remaining.

But Lukaku struck twice more in quick succession and even though Harry Arter got another back for Bournemouth, Ross Barkley grabbed a sixth goal for Everton in stoppage time.

Everton is seventh in the league, with Bournemouth 14th.

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