Solve by completing the squ \[ x^{2}+14 x-5=0 \] The solution set is

To solve the equation \( x^2 + 14x - 5 = 0 \) by completing the square, we first need to rearrange the equation. Start with the equation: \[ x^2 + 14x - 5 = 0 \] 1. Move the constant term to the other side: \[ x^2 + 14x = 5 \] 2. Now, to complete the square, we take half of the coefficient of \( x \) (which is 14), square it, and add it to both sides. Half of 14 is 7, and squaring it gives 49: \[ x^2 + 14x + 49 = 5 + 49 \] 3. This simplifies to: \[ (x + 7)^2 = 54 \] 4. Now take the square root of both sides: \[ x + 7 = \pm \sqrt{54} \] 5. Simplifying \( \sqrt{54} \), we have: \[ \sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6} \] 6. Thus, we have: \[ x + 7 = 3\sqrt{6} \quad \text{or} \quad x + 7 = -3\sqrt{6} \] 7. Finally, solving for \( x \): \[ x = -7 + 3\sqrt{6} \quad \text{or} \quad x = -7 - 3\sqrt{6} \] The solution set is: \[ \left\{ -7 + 3\sqrt{6}, -7 - 3\sqrt{6} \right\} \]