Selidiki sifat persamaan kuadrat tolong jawaban ini​

x² – x + 5 = 0

a = 1, b = -1, c = 5

D = b² – 4ac

= ( -1 )² – 4.1.5

= 1 – 20

= -19

Karena D < 0, maka akarnya tidak real

x² + 2x – 3 = 0

a = 1, b = 2, c = -3

D = b² – 4ac

= 2² – 4.1 ( -3 )

= 4 + 12

= 16

Karena D > 0, maka akarnya real dan berbeda

2x² + 3x – 9 = 0

a = 2, b = 3, c = -9

D = b² – 4ac

= 3² – 4.2 ( -9 )

= 9 + 72

= 81

Karena D > 0, maka akarnya real dan berbeda