Răspuns :
Răspuns:
x={[tex]\sqrt{3} -1 ; -5-\sqrt{3}[/tex]}
Explicație pas cu pas:
[tex]x^2+6x+7=5+4\sqrt{3}|-5+4\sqrt{4} \\x^2+6x+2-4\sqrt{3} =0\\\\x=\frac{-6+\sqrt{6^2-4*1(2-4\sqrt{3} }) }{2*1}\\\\x=\frac{-6+\sqrt{36-4(2-4\sqrt{3} }) }{2} \\\\x=\frac{-6+\sqrt{6-2}(2+\sqrt{3}) }{2} \\\\x=\frac{-6+2(2+\sqrt{3}) }{2}[/tex]
[tex]x=\frac{2\sqrt{3} -2 }{2} \\\\x=\frac{2(\sqrt{3}-1) }{2}\\x=\sqrt{3} -1[/tex]
[tex]x=\frac{-6-\sqrt{6^2-4*1(2-4\sqrt{3}) } }{2*1} \\x=\frac{-6-\sqrt{36-4(2-4\sqrt{3}) } }{2}[/tex]
[tex]x=\frac{-6-\sqrt{-4(2-4\sqrt{3})+36 } }{2} \\\\x=\frac{-10-2\sqrt{3} }{2} ,x=\frac{-2(5+\sqrt{3}) }{2} \\\\x=-(5+\sqrt{3} )\\x=-5-\sqrt{3} \\x=\sqrt{3} -1,x=-5-\sqrt{3}[/tex]
x={[tex]\sqrt{3}-1 ; -5-\sqrt{3}[/tex]}
Succes! :)