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Jika dibelakang y² terdapat angka yg lebih besar dari 1, Maka menggunakan rumus =
... * ... = ac
... + ... = b
... * ... = 2(-3)
... + ... = 5
6 * -1 = -6
6 + (-1) = 5
6 = p
-1 = q
Lalu, ..
a (y + p/a) (y + q/a)
2 (y + 6/2) (y + (-1)/2)
(y + 3) (2y -1)
Kalau ingin mencari y ..
y + 3 = 0
y = -3 ---> y1
2y - 1 = 0
y = 1/2 -> y2