29. Solve: $$ 3 x^{2}-4 x-3=0 $$

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$$
\textbf{Step 1: Identify the coefficients} 
$$
Given the quadratic equation,
$$
3x^2 - 4x - 3 = 0,
$$
we have:
$$
a = 3, \quad b = -4, \quad c = -3.
$$

$$
\textbf{Step 2: Use the quadratic formula} 
$$
The quadratic formula is:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$

$$
\textbf{Step 3: Calculate the discriminant} 
$$
The discriminant $$ \Delta $$ is given by:
$$
\Delta = b^2 - 4ac = (-4)^2 - 4(3)(-3) = 16 + 36 = 52.
$$

$$
\textbf{Step 4: Substitute back into the quadratic formula} 
$$
Substituting $$ a = 3 $$, $$ b = -4 $$, and $$ \Delta = 52 $$ into the quadratic formula, we get:
$$
x = \frac{-(-4) \pm \sqrt{52}}{2 \times 3} = \frac{4 \pm \sqrt{52}}{6}.
$$

$$
\textbf{Step 5: Simplify the square root} 
$$
Notice that:
$$
\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}.
$$
Thus, the expression for $$ x $$ becomes:
$$
x = \frac{4 \pm 2\sqrt{13}}{6}.
$$

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\textbf{Step 6: Simplify the fraction} 
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Divide both the numerator and the denominator by 2:
$$
x = \frac{2 \pm \sqrt{13}}{3}.
$$

$$
\textbf{Final Answer:}
$$
$$
x = \frac{2 + \sqrt{13}}{3} \quad \text{or} \quad x = \frac{2 - \sqrt{13}}{3}.
$$
The solutions are $$ x = \frac{2 + \sqrt{13}}{3} $$ and $$ x = \frac{2 - \sqrt{13}}{3} $$.

29. Solve: \[ 3 x^{2}-4 x-3=0 \]

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