Find the solution(s) using the quadratic formula.
$4{(x-2)}^2-5=x+7$

• A. $x=5,x=\frac{1}{2}$
• B. $x=5,x=\frac{1}{4}$
• C. $x=4,x=\frac{1}{4}$
• D. $x=4,x=\frac{1}{2}$

Answer: (C)

Verified step by step guidance

1. Start with the given equation: \(4\left(x - 2\right)^2 - 5 = x + 7\).
2. First, expand the squared term: \(\left(x - 2\right)^2 = x^2 - 4x + 4\). Substitute this back into the equation to get \$4(x^2 - 4x + 4) - 5 = x + 7$.
3. Distribute the 4 across the terms inside the parentheses: \$4x^2 - 16x + 16 - 5 = x + 7$.
4. Simplify the left side by combining like terms: \$4x^2 - 16x + 11 = x + 7\(. Then, move all terms to one side to set the equation equal to zero: \)4x^2 - 16x + 11 - x - 7 = 0\(, which simplifies to \)4x^2 - 17x + 4 = 0$.
5. Now, apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=4\), \(b=-17\), and \(c=4\). Substitute these values into the formula to find the solutions for \(x\).