Réponse :
Bonsoir
Explications étape par étape
x²-3<(4X+9)(X-3)
X^2-3< 4x^2 - 12x + 9x - 27
X^2 - 4x^2 +12x-9x < -27 + 3
-3x^2 + 3x < -24
-3x^2+3x+24<0
Delta = b^2-4ac
a=-3
b=3
c=24
D= (3)^2-4*(-3)*24
D=9+288
D=297
297>0 Donc 2 racines
X1 = -b-VDelta/2a
X1 = -3 - V297/2*(-3)
X1 = -3 - 3V33/-6
X1 = 3 - 3V33/6
X2 = -b+VDelta/2a
X2 = -3+V297/2*(-3)
X2 = -3 + 3V33/-6
X2 = 3+3V33/6
S = ]-infini ; 3-3V33/6[ U ]3+3V33/6 ; +infini[
