1. **State the problem:** Solve the quadratic equation $$x^2 - 14x + 38 = 0$$ by completing the square. 2. **Recall the formula and rule:** To complete the square for an equation of the form $$x^2 + bx + c = 0$$, we rewrite it as $$\left(x - \frac{b}{2}\right)^2 = \text{some number}$$ by adding and subtracting the square of half the coefficient of $x$. 3. **Rewrite the equation:** $$x^2 - 14x + 38 = 0$$ Move the constant term to the right side: $$x^2 - 14x = -38$$ 4. **Complete the square:** Take half of $-14$, which is $-7$, and square it: $$\left(\frac{-14}{2}\right)^2 = (-7)^2 = 49$$ Add and subtract 49 on the left side: $$x^2 - 14x + 49 - 49 = -38$$ Group the perfect square trinomial: $$\left(x^2 - 14x + 49\right) - 49 = -38$$ 5. **Simplify:** $$\left(x - 7\right)^2 - 49 = -38$$ Add 49 to both sides: $$\left(x - 7\right)^2 = -38 + 49$$ $$\left(x - 7\right)^2 = 11$$ 6. **Solve for $x$ by taking the square root:** $$x - 7 = \pm \sqrt{11}$$ 7. **Isolate $x$:** $$x = 7 \pm \sqrt{11}$$ **Final answer:** $$x = 7 + \sqrt{11} \quad \text{or} \quad x = 7 - \sqrt{11}$$