1. **State the problem:** We have a rectangle with surface area 60 cm², length $x+9$, and width $x-8$. We need to find $x$ and the dimensions. 2. **Formula:** Area of rectangle = length $\times$ width. 3. **Set up the equation:** $$ (x+9)(x-8) = 60 $$ 4. **Expand the product:** $$ x^2 - 8x + 9x - 72 = 60 $$ $$ x^2 + x - 72 = 60 $$ 5. **Bring all terms to one side:** $$ x^2 + x - 72 - 60 = 0 $$ $$ x^2 + x - 132 = 0 $$ 6. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=1$, $c=-132$. Calculate discriminant: $$ \Delta = 1^2 - 4 \times 1 \times (-132) = 1 + 528 = 529 $$ Square root: $$ \sqrt{529} = 23 $$ Find roots: $$ x = \frac{-1 \pm 23}{2} $$ Two solutions: $$ x_1 = \frac{-1 + 23}{2} = \frac{22}{2} = 11 $$ $$ x_2 = \frac{-1 - 23}{2} = \frac{-24}{2} = -12 $$ 7. **Check for valid dimensions:** Width = $x-8$, must be positive. For $x=11$, width = $11 - 8 = 3$ (valid). For $x=-12$, width = $-12 - 8 = -20$ (invalid). 8. **Final answer:** $$ x = 11 $$ Length = $11 + 9 = 20$ cm Width = $11 - 8 = 3$ cm The rectangle dimensions are 20 cm by 3 cm.