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