Solve the following equation using quadratic formula:

2x² + $\sqrt{7}x$ - 7 = 0

Comparing equation 2x² + $\sqrt{7}x$ - 7 = 0 with ax² + bx + c = 0, we get :

a = 2, b = $\sqrt{7}$ and c = -7.

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(\sqrt{7}) \pm \sqrt{(\sqrt{7})^2 - 4 \times 2 \times (-7)}}{2 \times 2} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{7 + 56}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{63}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm \sqrt{7 \times 9}}{4} \\[1em] = \dfrac{-\sqrt{7} \pm 3\sqrt{7}}{4} \\[1em] = \dfrac{-\sqrt{7} + 3\sqrt{7}}{4} \text{ or } \dfrac{-\sqrt{7} - 3\sqrt{7}}{4} \\[1em] = \dfrac{2\sqrt{7}}{4} \text{ or } \dfrac{-4\sqrt{7}}{4} \\[1em] = \dfrac{\sqrt{7}}{2} \text{ or } -\sqrt{7}.$

Hence, $x = \Big{-\sqrt{7}, \dfrac{\sqrt{7}}{2}\Big}$.