Solve for \( x \) using the quadratic formula. \( x^{2}-6 x+13=0 \)

To solve the quadratic equation \( x^{2} - 6x + 13 = 0 \) using the quadratic formula, we first identify the coefficients: \( a = 1 \), \( b = -6 \), and \( c = 13 \). The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Now, plug in the values: 1. Calculate \( b^2 - 4ac \): \[ b^2 - 4ac = (-6)^2 - 4(1)(13) = 36 - 52 = -16 \] 2. Since this results in a negative value, we will have complex roots. Now, compute \( x \): \[ x = \frac{-(-6) \pm \sqrt{-16}}{2(1)} = \frac{6 \pm 4i}{2} = 3 \pm 2i \] Thus, the solutions are: \[ x = 3 + 2i \quad \text{and} \quad x = 3 - 2i \]