Solve each quadratic equation using the square root property.
${(2z-3)}^2=16$

• A. $z=\pm \frac{7}{2}$
• B. $z=-\frac{1}{2}$ and $z=\frac{7}{2}$
• C. $z=-\frac{1}{2}$ and $z=\frac{2}{7}$
• D. $z=-\frac{2}{7}$ and $z=\frac{7}{2}$

Answer: (B)

Verified step by step guidance

1. Start with the given equation: \(\left(2z - 3\right)^2 = 16\).
2. Apply the square root property by taking the square root of both sides, remembering to include the plus-minus symbol: \$2z - 3 = \pm \sqrt{16}$.
3. Simplify the square root: \$2z - 3 = \pm 4$.
4. Set up two separate linear equations to solve for \(z\): 1) \$2z - 3 = 4\( 2) \)2z - 3 = -4$.
5. Solve each linear equation for \(z\) by isolating \(z\): For \$2z - 3 = 4\(, add 3 to both sides and then divide by 2. For \)2z - 3 = -4$, add 3 to both sides and then divide by 2.