$5 x^{2}-6 x+2=0$
Given: $5 x^{2}-6 x+2=0$
Comparing the given equation with general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=5, b=-6$ and $c=2$.
Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:
$\alpha=\frac{6+\sqrt{36-4 \times 5 \times 2}}{2 \times 5}$ and $\beta=\frac{6-\sqrt{36-4 \times 2 \times 5}}{2 \times 5}$
$\Rightarrow \alpha=\frac{6+\sqrt{-4}}{10} \quad$ and $\quad \beta=\frac{6-\sqrt{-4}}{10}$
$\Rightarrow \alpha=\frac{6+\sqrt{4 i^{2}}}{10} \quad$ and $\quad \beta=\frac{6-\sqrt{4 i^{2}}}{10}$
$\Rightarrow \alpha=\frac{6+2 i}{10} \quad$ and $\quad \beta=\frac{6-2 i}{10}$
$\Rightarrow \alpha=\frac{2(3+i)}{10}$ and $\quad \beta=\frac{2(3-i)}{10}$
$\Rightarrow \alpha=\frac{3}{5}+\frac{1}{5} i \quad$ and $\quad \beta=\frac{3}{5}-\frac{1}{5} i$
Hence, the roots of the equation are $\frac{3}{5} \pm \frac{1}{5} i$.