y=ax²+bx+c

x=-b/2a

Fais varier les coefficients a, b et c.
Quelle conjecture peux tu faire entre le signe  
de b²-4ac et le nombre de solutions de 
l'équation (E) : f(x) = 0 ?

a = 7 (slider from -10 to 10, step 1)
b = 9 (slider from -10 to 10, step 1)
c = 3 (slider from -10 to 10, step 1)
f(x) = (a * x^(2)) + (b * x) + c
ac: (a * c)
d: x = ((-b)) / ((2 * a))
X', X'' = Intersect(f, xAxis)
text3: "Δ=b²-4ac = (" + b + ")² - 4×(" + a + ")×(" + c + ") = " + (b^(2) - (4 * ac)) + ""
text1: "L'expression du trinôme est :
f(x) = " + a + " x² +" + b + " x +" + c + " "