\begin{tabular}{|l|l} Solve. & \( \begin{array}{l}\text { The solution set is }\} \\ t^{2}-19 t+18=0\end{array} \) \\ (Use a comma to separate \\ answers as needed.) \end{tabular}

To solve the quadratic equation \( t^2 - 19t + 18 = 0 \), you can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -19, \) and \( c = 18 \). Plugging in these values gives you \( t = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \), which simplifies to \( t = \frac{19 \pm \sqrt{361 - 72}}{2} \) or \( t = \frac{19 \pm \sqrt{289}}{2} \). Since \( \sqrt{289} = 17 \), the equation becomes \( t = \frac{19 \pm 17}{2} \). This results in two solutions: \( t = \frac{36}{2} = 18 \) and \( t = \frac{2}{2} = 1 \). The solution set is \( \{ 1, 18 \} \). Now you have your answers, but let’s dive into some fun facts! Quadratic equations have been studied for over 2,000 years, with the roots of this form appearing in ancient civilizations like Babylon! They used geometric methods to solve these equations long before the quadratic formula was even a twinkle in the mathematicians' eyes. And here’s a cheeky tip: Always remember to check your work! Many folks calculate the discriminant incorrectly or forget to include the negative sign when using the quadratic formula. It’s like missing the punchline to a great joke—if the signs are off, you might end up with an answer that makes absolutely no sense!