Find the positive difference between the solutions to the equation $6t^2 + 30 = 41t$.
To find the positive difference between the solutions, I need to first find the solutions themselves. I can use the quadratic formula to find the solutions of a quadratic equation of the form $at^2 + bt + c = 0$, where $a, b, c$ are constants. In this case, $a = 6, b = -41, c = 30$, so I can plug these values into the formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Simplifying, I get $t = \frac{41 \pm \sqrt{1681 - 720}}{12} = \frac{41 \pm \sqrt{961}}{12}$. Since $\sqrt{961} = 31$, I can write $t = \frac{41 \pm 31}{12}$. This gives me two solutions: $t = \frac{41 + 31}{12} = 6$ and $t = \frac{41 - 31}{12} = \frac{5}{6}$. The positive difference between these solutions is $|6 - \frac{5}{6}| = \frac{31}{6}$.

