1. **State the problem:** Rewrite the quadratic expression $x^2 + 10x + 29$ in the form $(x + c)^2 + d$, where $c$ and $d$ are integers. 2. **Formula and rule:** To complete the square for a quadratic expression $x^2 + bx + c$, use the formula: $$x^2 + bx + c = (x + \frac{b}{2})^2 + \left(c - \left(\frac{b}{2}\right)^2\right)$$ This means we take half of the coefficient of $x$, square it, and add and subtract that inside the expression. 3. **Apply to the problem:** Here, $b = 10$ and $c = 29$. Calculate $\frac{b}{2} = \frac{10}{2} = 5$. Square it: $5^2 = 25$. 4. **Rewrite the expression:** $$x^2 + 10x + 29 = (x + 5)^2 + (29 - 25) = (x + 5)^2 + 4$$ 5. **Values of $c$ and $d$:** $c = 5$, $d = 4$. 6. **Turning point coordinates:** The quadratic in vertex form is $y = (x + 5)^2 + 4$. The turning point (vertex) is at $x = -c = -5$, and $y = d = 4$. **Answer:** The turning point is at $(-5, 4)$.