Solve. \( \begin{array}{l}r(r-4)=32\end{array} \begin{array}{l}\text { The solution set is }\{\square\} . \\ \text { (Use a comma to } \\ \text { separate answers.) }\end{array} \)

We start with the equation: \[ r(r-4)=32 \] **Step 1: Expand the left-hand side** \[ r^2 - 4r = 32 \] **Step 2: Rearrange the equation into standard quadratic form** Subtract \(32\) from both sides: \[ r^2 - 4r - 32 = 0 \] **Step 3: Factor the quadratic** We look for two numbers that multiply to \(-32\) and add to \(-4\). These numbers are \(-8\) and \(4\) because: \[ -8 \times 4 = -32 \quad \text{and} \quad -8 + 4 = -4 \] Thus, we factor the quadratic as: \[ (r - 8)(r + 4) = 0 \] **Step 4: Solve for \(r\) by setting each factor equal to zero** Set \(r - 8 = 0\): \[ r = 8 \] Set \(r + 4 = 0\): \[ r = -4 \] **Final Answer:** The solution set is \(\{-4, 8\}\).