The given equation is

$\begin{array}{l}
x+\frac{1}{x}=3, x \neq 0 \\
\Rightarrow \frac{x^{2}+1}{x}=3 \\
\Rightarrow x^{2}+1=3 x \\
\Rightarrow x^{2}-3 x+1=0
\end{array}$

This equation is of the form $a x^{2}+b x+c=0$, where, $a=1, b=-3$ and $c=1$.

$\therefore$ Discriminant,
$D=b^{2}-4 a c=(-3)^{2}-4 \times 1 \times 1=9-4=5>0$

So, the given equation has real roots.
Now, $\sqrt{D}=\sqrt{5}$

$\begin{array}{l}
\therefore a=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-3)+\sqrt{5}}{2 \times 1}=\frac{3+\sqrt{5}}{2} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-3)-\sqrt{5}}{2 \times 1}=\frac{3-\sqrt{5}}{2}
\end{array}$

Hence, $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the roots of the given equation.