1. **State the problem:** Rewrite the quadratic expression $2x^2 + 20x + 28$ in the form $a(x + b)^2 + c$, where $a$, $b$, and $c$ are integers. 2. **Formula and method:** We use the method of completing the square for quadratics: $$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - a\left(\frac{b}{2a}\right)^2\right)$$ 3. **Apply to the given quadratic:** Given $2x^2 + 20x + 28$, here $a=2$, $b=20$, and $c=28$. 4. **Calculate $b/(2a)$:** $$\frac{20}{2 \times 2} = \frac{20}{4} = 5$$ 5. **Rewrite the quadratic:** $$2x^2 + 20x + 28 = 2\left(x^2 + 10x\right) + 28$$ 6. **Complete the square inside the parentheses:** Add and subtract $5^2 = 25$ inside the parentheses: $$2\left(x^2 + 10x + 25 - 25\right) + 28 = 2\left((x + 5)^2 - 25\right) + 28$$ 7. **Distribute and simplify:** $$2(x + 5)^2 - 2 \times 25 + 28 = 2(x + 5)^2 - 50 + 28 = 2(x + 5)^2 - 22$$ 8. **Identify $a$, $b$, and $c$:** $$a = 2, \quad b = 5, \quad c = -22$$ 9. **Find the turning point:** The turning point of $y = a(x + b)^2 + c$ is at $(-b, c)$. So, the turning point is: $$(-5, -22)$$ **Final answers:** - $a = 2$ - $b = 5$ - $c = -22$ - Turning point coordinates: $(-5, -22)$