1. **State the problem:** Solve the equation $$2x^4 + 6x^2 - 8 = 0$$ for $x$. 2. **Use substitution:** Let $y = x^2$. Then the equation becomes $$2y^2 + 6y - 8 = 0$$. 3. **Simplify the quadratic equation:** Divide the entire equation by 2 to make it simpler: $$y^2 + 3y - 4 = 0$$. 4. **Factor the quadratic:** Find two numbers that multiply to $-4$ and add to $3$. These are $4$ and $-1$. $$y^2 + 4y - y - 4 = 0$$ $$(y + 4)(y - 1) = 0$$. 5. **Solve for $y$:** Set each factor equal to zero: $$y + 4 = 0 \Rightarrow y = -4$$ $$y - 1 = 0 \Rightarrow y = 1$$. 6. **Recall substitution:** Since $y = x^2$, solve for $x$: - For $y = -4$, $x^2 = -4$ has no real solutions because the square of a real number cannot be negative. - For $y = 1$, $x^2 = 1$ gives $x = \pm 1$. 7. **Final answer:** The real solutions are $$x = \pm 1$$. **Note:** The solutions $x = \pm \sqrt{-4}$ are complex and not real numbers.