(x+1)² = (2x-3)²
<-> x²+2x+1 = 4x²-12x+9
<-> -3x²+2x+1 = -12x+9
<-> -3x²+14x+1 = +9
<-> -3x²+14x-8 = 0
Δ = b²-4ac
Δ = 14²-4*(-3)*(-8)
Δ = 100
Δ > 0 donc l'équation possède 2 solutions x1 et x2 :
x1 = (-b-√Δ)/2a
x1 = (-14-√100)/2*(-3)
x1 = -24/(-6)
x1 = 4
x2 = (-b+√Δ)/2a
x2 = (-14+√100)/2*(-3)
x2 = -4/(-6)
x2 = 2/3
L'équation a donc pour solutions S = {2/3;4}
