S obzirom na zadanu vrijednost s promjenom # 4 (1 + y) x ^ 2-4xy- (1-il) # # => 4 (1 + y) 2-2 x ^ (1 + y) x 2 + (1-il) x- (1-il) # # => 2 (1 + y) x (2 x-1) + (1-il) (2 x-1) # # => (2 x-1), (2 (1 + y) + x (1-il)) = 0 # Stoga # X = 1/2 # provjeravanje # 4 (1 + y) x ^ 2-4xy- (1-il) # # = 4 (1 + y) (1/2) ^ 2-4 (1/2) y- (1-il) # # = 1 + y-2y-1 + y = 0 #
