1. **State the problem:** Solve the inequality $2x^2 + 6 > 40$ to find the values of $x$ that satisfy it. 2. **Rewrite the inequality:** Subtract 40 from both sides to set the inequality to zero: $$2x^2 + 6 - 40 > 0$$ which simplifies to $$2x^2 - 34 > 0$$ 3. **Divide both sides by 2:** $$x^2 - 17 > 0$$ 4. **Rewrite the inequality:** $$x^2 > 17$$ 5. **Solve for $x$:** Since $x^2 > 17$, $x$ must be greater than $\\sqrt{17}$ or less than $-\\sqrt{17}$. 6. **Final solution:** $$x < -\\sqrt{17} \quad \text{or} \quad x > \\sqrt{17}$$ This means $x$ is any real number outside the interval $[-\\sqrt{17}, \\sqrt{17}]$. **Answer:** $x < -\\sqrt{17}$ or $x > \\sqrt{17}$