Solve the equation using the quadratic formula $$ 3 x^{2}=-12 x-4 $$

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Given the equation
$$
3x^{2} = -12x - 4,
$$
we first rewrite it in standard quadratic form $$ ax^{2} + bx + c = 0 $$.

1. **Rearrange the equation:**

Add $$12x + 4$$ to both sides:
   $$
   3x^{2} + 12x + 4 = 0.
   $$
   
   Here, $$ a = 3 $$, $$ b = 12 $$, and $$ c = 4 $$.

2. **Apply the quadratic formula:**

The quadratic formula is given by:
   $$
   x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.
   $$
   Substitute $$ a = 3 $$, $$ b = 12 $$, and $$ c = 4 $$ into the formula:

$$
   x = \frac{-12 \pm \sqrt{12^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3}.
   $$

3. **Simplify inside the square root:**

Calculate the discriminant:
   $$
   12^{2} - 4 \cdot 3 \cdot 4 = 144 - 48 = 96.
   $$
   So, the expression becomes:
   $$
   x = \frac{-12 \pm \sqrt{96}}{6}.
   $$

4. **Simplify the square root:**

Factor $$96$$ to simplify:
   $$
   \sqrt{96} = \sqrt{16 \cdot 6} = \sqrt{16} \cdot \sqrt{6} = 4\sqrt{6}.
   $$
   Substitute back into the formula:
   $$
   x = \frac{-12 \pm 4\sqrt{6}}{6}.
   $$

5. **Simplify the fraction:**

Divide the numerator and denominator by $$2$$:
   $$
   x = \frac{-12 \pm 4\sqrt{6}}{6} = \frac{-6 \pm 2\sqrt{6}}{3}.
   $$

Thus, the solutions to the equation are:
$$
x = \frac{-6 + 2\sqrt{6}}{3} \quad \text{and} \quad x = \frac{-6 - 2\sqrt{6}}{3}.
$$
The solutions are $$ x = \frac{-6 + 2\sqrt{6}}{3} $$ and $$ x = \frac{-6 - 2\sqrt{6}}{3} $$.

Solve the equation using the quadratic formula

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