Considerando l'eqution dato con un cambiamento
# 4 (1 + y) x ^ 2-4xy- (1-y) #
# => 4 (1 + y) x ^ 2-2 (1 + y) x + 2 (1-y) x- (1-y) #
# => 2 (1 + y) x (2x-1) + (1-y) (2x-1) #
# => (2x-1) (2 (1 + y) x + (1-y)) = 0 #
Quindi # X = 1/2 #
verifica
# 4 (1 + y) x ^ 2-4xy- (1-y) #
# = 4 (1 + y) (1/2) ^ 2-4 (1/2) y- (1-y) #
# = 1 + y-2y-1 + y = 0 #