## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions to the equation $(2x-5)^{2}=81$ are $x=-2$ and $x=7$
1. $(2x-5)^{2}=81$ 2. Take the square root of each side: $\sqrt(2x-5)^{2}=\sqrt81$:$2x-5=±9$ 3. Add 5 to each side: $2x=±9+5$ 4. Divide each side by 2: $x=\frac{±9+5}{2}$ 5. Solve: $x=\frac{14}{2}=7$ and $x=\frac{-4}{2}=-2$