6x² − 19x + 10 = 0 a = 6 , b = -19 et c = 10
Δ = b² − 4ac = (-19)² − 4×6×10 = 121
Δ > 0 alors l'équation 6x² − 19x + 10 = 0 admet 2 solutions x1 et x2
v121 = 11
x1 = (-b − vΔ)/2a = (19 − 11) / 12 = 2/3
x2 = (-b + vΔ)/2a = (19 + 11) / 12 = 5/2:
Factorisation: a(x-x1)(x-x2)⇒6(x − 2/3)(x − 5/2)
S= [2/3 ; 5/2]
