1. Problem: Solve the quadratic equation $3x^2 + 12 = 0$. 2. Formula and rules: A quadratic equation is generally in the form $ax^2 + bx + c = 0$. Here, $a=3$, $b=0$, and $c=12$. To solve, isolate $x^2$ and then take the square root. 3. Intermediate work: $$3x^2 + 12 = 0$$ $$3x^2 = -12$$ $$\cancel{3}x^2 = \cancel{3}(-4)$$ $$x^2 = -4$$ 4. Since $x^2 = -4$, and the square of a real number cannot be negative, there are no real solutions. The solutions are complex: $$x = \pm \sqrt{-4} = \pm 2i$$ 5. Explanation: We moved the constant to the other side, divided both sides by 3, and found that $x^2$ equals a negative number, which means the solutions are imaginary numbers $\pm 2i$. Final answer: $x = \pm 2i$