Solve for all solutions of \( x \) : \( \begin{array}{l}4 x^{2}-32 x+80=0 \\ x=6,-2 \\ x=4 \pm 2 i \\ x=-4 \pm 2 i \\ x=4 \pm i \sqrt{2}\end{array} \)

To solve the equation \( 4x^2 - 32x + 80 = 0 \), we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -32 \), and \( c = 80 \). Calculating the discriminant: \[ b^2 - 4ac = (-32)^2 - 4 \cdot 4 \cdot 80 = 1024 - 1280 = -256 \] Since the discriminant is negative, we will have complex solutions. Now, substituting \( a \), \( b \), and the calculated discriminant into the quadratic formula gives: \[ x = \frac{32 \pm \sqrt{-256}}{8} = \frac{32 \pm 16i}{8} = 4 \pm 2i \] Thus, the solutions to the equation are: \[ x = 4 + 2i \quad \text{and} \quad x = 4 - 2i \] Therefore, the correct solutions for \( x \) from the options provided are \( x = 4 \pm 2i \).