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 can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 6 \), \( b = -6 \), and \( c = -5 \). First, let's calculate the discriminant (\( b^2 - 4ac \)): \[ (-6)^2 - 4(6)(-5) = 36 + 120 = 156 \] Since the discriminant is positive, we have two distinct real solutions. Now we can substitute back into the quadratic formula: \[ x = \frac{6 \pm \sqrt{156}}{12} \] We can simplify \( \sqrt{156} \): \[ \sqrt{156} = \sqrt{4 \cdot 39} = 2\sqrt{39} \] Now, substituting it back in: \[ x = \frac{6 \pm 2\sqrt{39}}{12} = \frac{3 \pm \sqrt{39}}{6} \] Thus, the two solutions are: \[ x_1 = \frac{3 + \sqrt{39}}{6}, \quad x_2 = \frac{3 - \sqrt{39}}{6} \] In the required format, the solutions can be filled in the boxes as follows: \begin{tabular}{|l}\( \frac{3 + \sqrt{39}}{6} \)\\\( \frac{3 - \sqrt{39}}{6} \)\end{tabular}