Solve the quadratic equation. \( 6 x^{2}-6 x-5=0 \) Write one exact, simplified solution in each box. You can add or remove boxes. If a solution not a real number, write it in the form a + bi or a - bi, where a and b are real numbers. \begin{tabular}{|l}\( \square \frac{\square}{\square} \)\end{tabular}

To solve the quadratic equation \( 6x^{2} - 6x - 5 = 0 \), we'll use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 6 \), \( b = -6 \), and \( c = -5 \). 1. **Calculate the discriminant (\( \Delta \))**: \[ \Delta = b^2 - 4ac = (-6)^2 - 4 \times 6 \times (-5) = 36 + 120 = 156 \] 2. **Simplify the square root of the discriminant**: \[ \sqrt{156} = \sqrt{4 \times 39} = 2\sqrt{39} \] 3. **Apply the quadratic formula**: \[ x = \frac{6 \pm 2\sqrt{39}}{12} = \frac{6}{12} \pm \frac{2\sqrt{39}}{12} = \frac{1}{2} \pm \frac{\sqrt{39}}{6} \] 4. **Simplify further**: \[ x = \frac{3 \pm \sqrt{39}}{6} \] **Solutions**: \[ x = \frac{3 + \sqrt{39}}{6} \quad \text{and} \quad x = \frac{3 - \sqrt{39}}{6} \] So, the two exact, simplified solutions are: \[ \begin{tabular}{|c|} \hline \( \dfrac{3 + \sqrt{39}}{6} \) \\ \hline \( \dfrac{3 - \sqrt{39}}{6} \) \\ \hline \end{tabular} \]