Algebra worksheet for solving quadratic equations, including writing the quadratic formula and solving equations by factoring or using the quadratic formula.

Problem Overview:

The task involves solving quadratic equations using one of three methods: factoring, completing the square, or the quadratic formula. The solutions should be written in their simplest form, and answers involving radicals should be left in exact form.

Quadratic Formula Recap:

The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where the quadratic equation is in the standard form:
\[
ax^2 + bx + c = 0
\]

Solving Each Equation:

#### 1. \( x^2 = 64 \)
This equation is already in a simple form where we can solve for \( x \) directly by taking the square root of both sides.

\[
x^2 = 64
\]
\[
x = \pm \sqrt{64}
\]
\[
x = \pm 8
\]

So, the solutions are:
\[
x = 8 \quad \text{and} \quad x = -8
\]

#### 2. \( x^2 + 6x = 0 \)
This equation can be solved by factoring. Notice that \( x \) is a common factor.

\[
x^2 + 6x = 0
\]
\[
x(x + 6) = 0
\]

Using the zero-product property:
\[
x = 0 \quad \text{or} \quad x + 6 = 0
\]
\[
x = 0 \quad \text{or} \quad x = -6
\]

So, the solutions are:
\[
x = 0 \quad \text{and} \quad x = -6
\]

#### 3. \( x^2 + 5x + 6 = 0 \)
This equation can be solved by factoring. We need to find two numbers that multiply to \( 6 \) (the constant term) and add up to \( 5 \) (the coefficient of \( x \)).

The numbers are \( 2 \) and \( 3 \):
\[
x^2 + 5x + 6 = (x + 2)(x + 3) = 0
\]

Using the zero-product property:
\[
x + 2 = 0 \quad \text{or} \quad x + 3 = 0
\]
\[
x = -2 \quad \text{or} \quad x = -3
\]

So, the solutions are:
\[
x = -2 \quad \text{and} \quad x = -3
\]

#### 4. \( 4x^2 + 2x + 3 = 0 \)
This equation does not factor easily, so we use the quadratic formula. Here, \( a = 4 \), \( b = 2 \), and \( c = 3 \).

The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values:
\[
x = \frac{-2 \pm \sqrt{2^2 - 4(4)(3)}}{2(4)}
\]
\[
x = \frac{-2 \pm \sqrt{4 - 48}}{8}
\]
\[
x = \frac{-2 \pm \sqrt{-44}}{8}
\]
\[
x = \frac{-2 \pm \sqrt{4 \cdot (-11)}}{8}
\]
\[
x = \frac{-2 \pm 2i\sqrt{11}}{8}
\]
\[
x = \frac{-1 \pm i\sqrt{11}}{4}
\]

So, the solutions are:
\[
x = \frac{-1 + i\sqrt{11}}{4} \quad \text{and} \quad x = \frac{-1 - i\sqrt{11}}{4}
\]

#### 5. \( x^2 - 4x = 1 \)
First, rewrite the equation in standard form:
\[
x^2 - 4x - 1 = 0
\]

Now, use the quadratic formula. Here, \( a = 1 \), \( b = -4 \), and \( c = -1 \).

The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values:
\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)}
\]
\[
x = \frac{4 \pm \sqrt{16 + 4}}{2}
\]
\[
x = \frac{4 \pm \sqrt{20}}{2}
\]
\[
x = \frac{4 \pm 2\sqrt{5}}{2}
\]
\[
x = 2 \pm \sqrt{5}
\]

So, the solutions are:
\[
x = 2 + \sqrt{5} \quad \text{and} \quad x = 2 - \sqrt{5}
\]

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad x = 8, \, x = -8 \\
2. & \quad x = 0, \, x = -6 \\
3. & \quad x = -2, \, x = -3 \\
4. & \quad x = \frac{-1 + i\sqrt{11}}{4}, \, x = \frac{-1 - i\sqrt{11}}{4} \\
5. & \quad x = 2 + \sqrt{5}, \, x = 2 - \sqrt{5}
\end{aligned}
}
\]