Answer :
The zeros of f(x) are the values of x for whixh f(x) = 0.
That is,
[tex]x^2-6x+3=0[/tex]Using the formula method, we have:
[tex]\begin{gathered} a=1,\text{ b=-6, c = 3} \\ x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(1)(3)}}{2(1)} \\ \Rightarrow x=\frac{6\pm\sqrt[]{36-12}}{2}=\frac{6\pm\sqrt[]{24}}{2}=\frac{6\pm2\sqrt[]{6}}{2}=3\pm\sqrt[]{6} \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow x=3\text{ + 2.45 or x = 3-2.45} \\ x\text{ =5.45 or 0.55} \end{gathered}[/tex]