Bonsoir,
x²-5x+6 = 0 Δ = 1 x ' = 2 et x " = 3
x²-4x+4 = 0 Δ = 0 un seule solution x = (-b)/2a = 2
x² - x - 1 = 0 Δ = 5 x ' = (1-√5)/2 et x" = (1+√5)/2
2x² - 8x - 10 = 0 Δ = 144 x ' = -1 et x " = 5
5x²+2x+1 Δ = -16 donc aucune solution car Δ < 0
(3/2)x² - 4x + 2.5 = 0 Δ = 1 x ' = 1 et x" = 5/3
7x² + 14x + 2 = 0 Δ = 140
x ' = (-14 - √140)/14 = (2(-7) - √ 4 *√ 35) / 2 * 7)
x' = (-7 -√35)/7
pareil pour x " = (-14 + √140)/14 = (-7 + √35)/7
Bonne soirée
