Solve the following equations by using formula:

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

The given equation is $2x^2 + \sqrt{5}x - 5 = 0$

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = $\sqrt{5}$ , c = -5

By using formula,

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

we obtain:

$\Rightarrow x = \dfrac{-(\sqrt{5}) ± \sqrt{(\sqrt{5})^2 - 4\times 2 \times -5}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{5 + 40}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} ± \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + \sqrt{45}}{4} \text{ or } \dfrac{-\sqrt{5} - \sqrt{45}}{4} \\[1em] \Rightarrow x = \dfrac{-\sqrt{5} + 3\sqrt{5}}{4} \text { or } \dfrac{-\sqrt{5} - 3\sqrt{5}}{4} \\[1em] \Rightarrow x = \dfrac{\sqrt{5}(-1 + 3)}{4} \text{ or } \dfrac{\sqrt{5}(-1 - 3)}{4} \\[1em] x = \dfrac{\sqrt{5}}{2} \text{ or } -\sqrt{5}$

Hence roots of the given equations are $\dfrac{\sqrt{5}}{2} , -\sqrt{5}$.