1. **State the problem:** Solve the quadratic equation $$\sqrt{2} x^2 + (\sqrt{2} - 2)x - 2 = 0$$ for $x$. 2. **Recall the quadratic formula:** For an equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = \sqrt{2}$, $b = \sqrt{2} - 2$, and $c = -2$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (\sqrt{2} - 2)^2 - 4 \times \sqrt{2} \times (-2)$$ 4. **Expand and simplify:** $$(\sqrt{2} - 2)^2 = (\sqrt{2})^2 - 2 \times 2 \times \sqrt{2} + 2^2 = 2 - 4\sqrt{2} + 4 = 6 - 4\sqrt{2}$$ 5. **Calculate $-4ac$ term:** $$-4 \times \sqrt{2} \times (-2) = 8\sqrt{2}$$ 6. **Sum the discriminant:** $$\Delta = 6 - 4\sqrt{2} + 8\sqrt{2} = 6 + 4\sqrt{2}$$ 7. **Apply the quadratic formula:** $$x = \frac{-(\sqrt{2} - 2) \pm \sqrt{6 + 4\sqrt{2}}}{2\sqrt{2}} = \frac{-\sqrt{2} + 2 \pm \sqrt{6 + 4\sqrt{2}}}{2\sqrt{2}}$$ 8. **Simplify numerator and denominator if possible:** No common factors to cancel, but we can write the solution as is. 9. **Final answer:** $$x = \frac{2 - \sqrt{2} \pm \sqrt{6 + 4\sqrt{2}}}{2\sqrt{2}}$$ This gives two real roots for the quadratic equation.