1. **State the problem:** Solve the quadratic equation $$3x^2 = -4 + 8x$$ for $x$. 2. **Rewrite the equation in standard form:** Move all terms to one side to get $$3x^2 - 8x + 4 = 0$$. 3. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 4. **Identify coefficients:** Here, $a=3$, $b=-8$, and $c=4$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-8)^2 - 4 \times 3 \times 4 = 64 - 48 = 16$$. 6. **Compute the square root of the discriminant:** $$\sqrt{16} = 4$$. 7. **Apply the quadratic formula:** $$x = \frac{-(-8) \pm 4}{2 \times 3} = \frac{8 \pm 4}{6}$$. 8. **Find the two solutions:** - For the plus sign: $$x = \frac{8 + 4}{6} = \frac{12}{6} = 2$$. - For the minus sign: $$x = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}$$. 9. **Final answer:** The solutions to the equation are $$x = 2$$ and $$x = \frac{2}{3}$$.