[tex]\displaystyle\bf\\x_{1} =3+\sqrt{5} \\\\x_{2} =3-\sqrt{5} \\\\Teorema \ Vieta:\\\\x^{2} +px+q=0\\\\x_{1} +x_{2} =-p\\\\x_{1} \cdot x_{2} =q\\\\-p=3+\sqrt{5} +3-\sqrt{5} =6\\\\p=-6\\\\q=(3+\sqrt{5} )\cdot(3-\sqrt{5} )=3^{2} -(\sqrt{5} )^{2} =9-5=4\\\\\\\boxed{\boxed{x^{2} -6x+4=0}}[/tex]