Solve the following

Given: $\sqrt{5} x^{2}+x+\sqrt{5}=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{5}, b=1$ and $c=\sqrt{5}$.

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:

$\alpha=\frac{-1+\sqrt{1-4 \times \sqrt{5} \times \sqrt{5}}}{2 \sqrt{5}}$ and $\quad \beta=\frac{-1-\sqrt{1-4 \times \sqrt{5} \times \sqrt{5}}}{2 \sqrt{5}}$

$\alpha=\frac{-1+\sqrt{-19}}{2 \sqrt{5}} \quad$ and $\quad \beta=\frac{-1-\sqrt{-19}}{2 \sqrt{5}}$

$\alpha=\frac{-1+i \sqrt{19}}{2 \sqrt{5}} \quad$ and $\quad \beta=\frac{-1-i \sqrt{19}}{2 \sqrt{5}}$

Hence, the roots of the equation are $\frac{-1 \pm i \sqrt{19}}{2 \sqrt{5}}$.