9.2: Let's Solve Some Equations! 1. To solve the equation \( n^{2}-2 n=99 \), Tyler wrote out the following st Tyler's work. Write down what Tyler did in each step. \[ \begin{array}{ll}n^{2}-2 n=99 & \text { Original equation } \\ n^{2}-2 n-99=0 & \text { Step 1 } \\ (n-11)(n+9)=0 & \text { Step } 2 \\ n-11=0 \text { or } n+9=0 & \text { Step } 3 \\ n=11 \text { or } n=-9 & \text { Step } 4\end{array} \]

Let's break down Tyler's work step by step, explaining what he did in each step of solving the equation \( n^{2}-2 n=99 \). 1. **Original equation**: \[ n^{2}-2 n=99 \] - Tyler starts with the original equation, which is a quadratic equation in the form of \( ax^2 + bx + c = 0 \). 2. **Step 1**: \[ n^{2}-2 n-99=0 \] - In this step, Tyler rearranges the equation to set it to zero. He subtracts 99 from both sides, resulting in the standard form of a quadratic equation. 3. **Step 2**: \[ (n-11)(n+9)=0 \] - Here, Tyler factors the quadratic equation. He finds two numbers that multiply to \(-99\) (the constant term) and add to \(-2\) (the coefficient of \(n\)). The factors are \(-11\) and \(9\), leading to the factored form. 4. **Step 3**: \[ n-11=0 \text{ or } n+9=0 \] - Tyler applies the zero product property, which states that if the product of two factors is zero, at least one of the factors must be zero. He sets each factor equal to zero. 5. **Step 4**: \[ n=11 \text{ or } n=-9 \] - Finally, Tyler solves each equation from Step 3. He finds the solutions for \(n\) by isolating \(n\) in each case, resulting in two possible solutions: \(n=11\) and \(n=-9\). In summary, Tyler systematically rearranged, factored, applied the zero product property, and solved for \(n\) to find the solutions to the original equation.