Bonjour,
[tex]x {}^{2} + 3x + 2 = 6x - 2[/tex]
[tex] {-x}^{2} - 3x + 4 = 0[/tex]
A= -1 B = -3 C = 4
∆=
[tex]b {}^{2} - 4ac[/tex]
[tex] = - 3 {}^{2} - 4 \times -1 \times 4[/tex]
[tex] = 9 +16[/tex]
[tex] = 25[/tex]
∆ est positif donc deux solutions
X1 = (-b +✓∆)/ 2a
=(3 + 5)/-2
= 8/-2
=-4
X2 = (-b - ✓∆)/2a
= (3 - 5)/-2
= -2/-2
= 1
S{-4;1}