[tex]x^2-4x+3=x+2\\ x^2 - 5x + 1 = 0\\ (x - \frac{5}{2})^2 - \frac{25}{4} + 1 = 0\\ (x - \frac{5}{2})^2 - \frac{25}{4} + \frac{4}{4} = 0\\ (x - \frac{5}{2})^2 - \frac{21}{4}} = 0\\ (x - \frac{5}{2})^2 - (\frac{ \sqrt{21}}{2})^2 = 0\\ (x - \frac{5}{2}- \frac{\sqrt{21}}{2})(x - \frac{5}{2}+ \frac{\sqrt{21}}{2}) = 0\\ donc\, 2\, solutions\,: \\ x= \frac{5}{2}+ \frac{\sqrt{21}}{2}\\ ou \\x= \frac{5}{2}- \frac{\sqrt{21}}{2}[/tex]
En espérant t'avoir aidé.
