developer 5=(x-3)*(2x+3) pour trouver x

5 = x * 2x + x * 3 - 3 * 2x - 3 * 3
5 = 2x^2 + 3x - 6x - 9
2x^2 - 3x - 9 - 5 = 0
2x^2 - 3x - 14 = 0

On calcule le discriminant :

Delta = (-3)^2 - 4 * 2 * (-14)
Delta = 9 + 112
Delta = 121 > 0 donc deux solutions
V(delta) = V121 = 11

X1 = (3 - 11) / (2 * 2) = (-8) / 4 = (-2)

X2 = (3 + 11) / (2 * 2) = 14 / 4 = 7/2

5=(x-3)*(2x+3)
5=(2x²-6x+3x -9)
2x²-3x-9-5=0
2x²-3x-14=0

On calcule le delta:

Le discriminant Δ = b² - 4ac = (-3)² − 4×2×-14 = 121 Δ > 0
alors l'équation 2x² − 3x − 14 = 0 admet 2 solutions réelles x1 et x2

On remarque que √121 = 11

x1 = (-b − √Δ)/2a = (3 − 11) / 4 = -2 

et
 x2 = (-b + √Δ)/2a = (3 + 11) / 4 = 7/2

 2x² − 3x − 14 comme factorisation : 2(x + 2)(x − 7/2)