1. **State the problem:** Solve the quadratic equation $$3x^2 + 7x - 4 = 0$$ using the completing the square method. 2. **Rewrite the equation:** Divide the entire equation by 3 to make the coefficient of $$x^2$$ equal to 1: $$x^2 + \frac{7}{3}x - \frac{4}{3} = 0$$ 3. **Isolate the constant term:** Move the constant term to the right side: $$x^2 + \frac{7}{3}x = \frac{4}{3}$$ 4. **Complete the square:** Take half of the coefficient of $$x$$, square it, and add to both sides. Half of $$\frac{7}{3}$$ is $$\frac{7}{6}$$. Square it: $$\left(\frac{7}{6}\right)^2 = \frac{49}{36}$$. Add $$\frac{49}{36}$$ to both sides: $$x^2 + \frac{7}{3}x + \frac{49}{36} = \frac{4}{3} + \frac{49}{36}$$ 5. **Simplify the right side:** Convert $$\frac{4}{3}$$ to $$\frac{48}{36}$$ to add easily: $$\frac{48}{36} + \frac{49}{36} = \frac{97}{36}$$ 6. **Write the left side as a perfect square:** $$\left(x + \frac{7}{6}\right)^2 = \frac{97}{36}$$ 7. **Solve for $$x$$:** Take the square root of both sides: $$x + \frac{7}{6} = \pm \sqrt{\frac{97}{36}} = \pm \frac{\sqrt{97}}{6}$$ 8. **Isolate $$x$$:** $$x = -\frac{7}{6} \pm \frac{\sqrt{97}}{6} = \frac{-7 \pm \sqrt{97}}{6}$$ **Final answer:** $$x = \frac{-7 + \sqrt{97}}{6} \quad \text{or} \quad x = \frac{-7 - \sqrt{97}}{6}$$