Boyd Holbrook deserves an award for his phenomenal performance as Clement Mansell in Justified: City Primeval! β€οΈππ₯
31.10.2023 05:34 β π 2 π 2 π¬ 1 π 0@chieko0901.bsky.social
Hi everyone! I love Narcos and Boyd Holbrook and Pedro Pascal and music and movies and drama!
Boyd Holbrook deserves an award for his phenomenal performance as Clement Mansell in Justified: City Primeval! β€οΈππ₯
31.10.2023 05:34 β π 2 π 2 π¬ 1 π 0Boyd Holbrook as Clement Mansell β€οΈ
Simply an appreciation post π
#BoydHolbrook
And Boydβs absolutely outstanding and phenomenal performance I mean. Heβs such a talent, really. He portrayed Mo to an absolute perfection. Itβs something special. He deserves/deserved an award for this as well. Love you and admire you, Boyd Holbrook! β€οΈπ
28.09.2023 00:51 β π 3 π 1 π¬ 0 π 0Boyd Holbrook at the Fan Expo Chicago, San Diego Comic Con and Vengeance premieres and screenings in summer last year.
You see Boydoβs Clement Mansell look?
Clement Mansell era.
Obviously because Boydo was shooting Justified that time (Summer 2022) β€οΈ
βThe Bikeridersβ will no longer be released on December 1 due to the ongoing SAG-AFTRA strike.
βThe hope is that βThe Bikeridersβ will succeed better β in terms of both box office and awards β once actors are able to promote the movie.β www.indiewire.com/news/general...
βThe Bikeridersβ to get special screening at 2023 AFI FEST on October 27 fest.afi.com/2023/special...
28.09.2023 21:27 β π 3 π 3 π¬ 0 π 0New/old photos of Boyd Holbrook (featuring Caroline Ribeiro and Evan Kidd) photographed by Steven Meisel for Vogue Italia, September 2002 instagram.com/p/CxbttBbum3A/
27.09.2023 21:12 β π 5 π 2 π¬ 0 π 1Hi everyone! This is my first post here so I'll introduce myself. I love Narcos, Boyd Holbrook and Pedro Pascal! And I love music, movies and dramas! I post on Twitter and Instagram (have accounts on Tumblr and Pinterest)My English is not very good. Please be my friendβΊοΈβ€οΈ
03.11.2023 02:24 β π 1 π 2 π¬ 0 π 0πβ€οΈ
03.11.2023 01:57 β π 1 π 0 π¬ 0 π 0π₯°β€οΈ
03.11.2023 01:56 β π 1 π 0 π¬ 0 π 0