1. **State the problem:** Solve the quadratic equation $8x^2 + 16x = 42$ by completing the square. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$8x^2 + 16x - 42 = 0$$ 3. **Divide through by the coefficient of $x^2$ to simplify:** $$\frac{8x^2}{8} + \frac{16x}{8} - \frac{42}{8} = 0$$ $$x^2 + 2x - \frac{21}{4} = 0$$ 4. **Isolate the constant term:** $$x^2 + 2x = \frac{21}{4}$$ 5. **Complete the square:** Take half the coefficient of $x$, which is $2$, half is $1$, then square it: $1^2 = 1$. Add $1$ to both sides: $$x^2 + 2x + 1 = \frac{21}{4} + 1$$ 6. **Simplify the right side:** $$\frac{21}{4} + \frac{4}{4} = \frac{25}{4}$$ 7. **Rewrite the left side as a perfect square:** $$(x + 1)^2 = \frac{25}{4}$$ 8. **Take the square root of both sides:** $$x + 1 = \pm \sqrt{\frac{25}{4}}$$ $$x + 1 = \pm \frac{5}{2}$$ 9. **Solve for $x$:** $$x = -1 \pm \frac{5}{2}$$ 10. **Find the two solutions:** $$x = -1 + \frac{5}{2} = \frac{3}{2}$$ $$x = -1 - \frac{5}{2} = -\frac{7}{2}$$ **Final answer:** $$x = \frac{3}{2} \text{ or } x = -\frac{7}{2}$$