Let's solve the equations step by step.

1. $x^2 + 14x + 49 = 4$

1. Subtract $4$ from both sides to set the equation equal to zero: $x^2 + 14x + 49 - 4 = 0$ $x^2 + 14x + 45 = 0$

2. Factor the quadratic: $x^2 + 14x + 45 = (x + 9)(x + 5)$

3. Solve for $x$ by setting each factor to zero: $x + 9 = 0 \quad \text{or} \quad x + 5 = 0$ $x = -9 \quad \text{or} \quad x = -5$

Solution for 1:

$x = -9 \quad \text{or} \quad x = -5$

2. $x^2 - 6x + 9 = 20$

1. Subtract $20$ from both sides: $x^2 - 6x + 9 - 20 = 0$ $x^2 - 6x - 11 = 0$

2. Factor or use the quadratic formula. This does not factor neatly, so use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -6$, $c = -11$.

   Substitute into the formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-11)}}{2(1)}$ $x = \frac{6 \pm \sqrt{36 + 44}}{2}$ $x = \frac{6 \pm \sqrt{80}}{2}$ Simplify $\sqrt{80}$: $\sqrt{80} = 4\sqrt{5}$ So: $x = \frac{6 \pm 4\sqrt{5}}{2}$ Simplify further: $x = 3 \pm 2\sqrt{5}$

Solution for 2:

$x = 3 + 2\sqrt{5} \quad \text{or} \quad x = 3 - 2\sqrt{5}$

Would you like further details or help with a specific step? Here are related questions for practice:

1. How do you determine if a quadratic is factorable?
2. What is the discriminant, and how does it help in solving quadratics?
3. Can you convert $x = 3 \pm 2\sqrt{5}$ into a decimal approximation?
4. What are perfect square trinomials, and how do you recognize them?
5. Can you rewrite the quadratic formula derivation step by step?

Tip: Always check if a quadratic can be factored easily before using the quadratic formula! It saves time and simplifies the process.