Answer:
[tex]\frac{3+ \sqrt{6} }{3}[/tex] and [tex]\frac{3- \sqrt{6} }{3}[/tex]
Step-by-step explanation:
This is a quadratic equation, so we start by rearranging its terms so we can apply the quadratic formula:
[tex]ax^2+bx+c=0[/tex] has two possible solutions given by:
[tex]x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}[/tex]
So we rearrange the terms in the given quadratic equation bringing all terms to the left hhand side of the equal sign:
[tex]3x^2+1=6x\\3x^2-6x+1=0[/tex]
The solutions will then be:
[tex]x=\frac{6+/- \sqrt{36-4(3)(1)} }{2(3)}=\frac{6+/- \sqrt{36-12} }{6}=\frac{6+/- \sqrt{24} }{6}=\frac{6+/- \sqrt{4*6} }{6}=\frac{6+/- 2\sqrt{6} }{6}=\frac{3+/- \sqrt{6} }{3}[/tex]
Therefore the two possible answer are: [tex]\frac{3+ \sqrt{6} }{3}[/tex] and [tex]\frac{3- \sqrt{6} }{3}[/tex]
