[tex](3x+5)^2-3^2 = 0\\(3x+5)^2=3^2\\3x+5=3\\3x=-2\\x=-\frac{2}{3}[/tex]
[tex](3x+5)^2-3^2=-5\\(3x+5)^2-9+5=0\\(3x+5)^2-4=0\\(3x+5)^2+2^2=0\\(3x+5+2)(3x+5-2)=0\\(3x+7)(3x+3)=0\\3x+7=0\\x=- \frac{7}{3}\\ou\\3x+3=0\\x=-1 [/tex]
En espérant t'avoir aidé.
