bonjour,
x²-4x-5 =0
(x-2)²-9 = 0
(x-2-3)(x-2+3) =0
(x-5)(x+1) = 0
x 5;-1
2x²+4x+12=0
pas de solution ds R
-4x²+2x=0
2x(-2x+1) =0
-2x+1 =0
-2x=-1
x = 1/2
2x =0
x = 0
x²+4x+2=0
(x+2)²-2 = 0 forme canonique
(x+2-√2)(x+2+√2) =0
x =-2+√2; -2-√2
25x²-20x+3=0
(5x-2)²-1 = 0 forme canonique
(5x-2-1)(5x-2+1) =0
(5x-3)(5x-1) =0
x =3/5 ; 1/5