1. **State the problem:** Express the quadratic expression $4x^2 - 12x + 13$ in the form $(2x + a)^2 + b$, where $a$ and $b$ are constants. 2. **Recall the formula:** To write a quadratic in the form $(mx + a)^2 + b$, we complete the square. The general approach is: $$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x\right) + c$$ Then complete the square inside the parentheses. 3. **Apply to our problem:** Given $4x^2 - 12x + 13$, factor out 4 from the first two terms: $$4x^2 - 12x + 13 = 4\left(x^2 - 3x\right) + 13$$ 4. **Complete the square inside the parentheses:** Take half of the coefficient of $x$, which is $-3$, so half is $-\frac{3}{2}$. Square it: $$\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$$ Add and subtract this inside the parentheses: $$4\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) + 13 = 4\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 13$$ 5. **Simplify:** Distribute 4: $$4\left(x - \frac{3}{2}\right)^2 - 4 \times \frac{9}{4} + 13 = 4\left(x - \frac{3}{2}\right)^2 - 9 + 13$$ Combine constants: $$4\left(x - \frac{3}{2}\right)^2 + 4$$ 6. **Rewrite in the desired form:** Since $4\left(x - \frac{3}{2}\right)^2 = (2x - 3)^2$, the expression is: $$(2x - 3)^2 + 4$$ **Final answer:** $$4x^2 - 12x + 13 = (2x - 3)^2 + 4$$