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

• A. $x=5,x=\(\frac\)12$
• B. $x=5,x=\(\frac\)14$
• C. $x=4,x=\(\frac\)14$
• D. $x=4,x=\(\frac\)12$

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 trinomial: \(4x^2 - 16x + 16 - 5 = x + 7\). Simplify the left side to \(4x^2 - 16x + 11 = x + 7\).
4. Bring 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. Identify coefficients for the quadratic formula: \(a = 4\), \(b = -17\), and \(c = 4\). Then apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions.