Let's solve the equations from your homework step by step:

1. Solve each equation.

(a) $0 = x^2 - 7x - 18$

We solve this quadratic equation by factoring. The equation can be rewritten as:

$x^2 - 7x - 18 = 0$

We need to find two numbers that multiply to $-18$ and add up to $-7$. These numbers are $-9$ and $2$. Thus, we can factor the quadratic as:

$(x - 9)(x + 2) = 0$

So, the solutions are:

$x = 9 \quad \text{or} \quad x = -2$

(b) $x^2 + 10x = 39$

Move 39 to the left side:

$x^2 + 10x - 39 = 0$

This quadratic doesn't factor easily, so we'll use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 10$, and $c = -39$. Plugging in the values:

$x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-39)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 156}}{2} = \frac{-10 \pm \sqrt{256}}{2}$

$x = \frac{-10 \pm 16}{2}$

So, the solutions are:

$x = \frac{-10 + 16}{2} = 3 \quad \text{or} \quad x = \frac{-10 - 16}{2} = -13$

(c) $0 = x^2 - 10x + 12$

This quadratic can be factored. We are looking for two numbers that multiply to 12 and add up to $-10$, which are $-6$ and $-4$:

$(x - 6)(x - 4) = 0$

So, the solutions are:

$x = 6 \quad \text{or} \quad x = 4$

(d) $2x^2 + 4x = 0$

Factor out $2x$:

$2x(x + 2) = 0$

So, the solutions are:

$x = 0 \quad \text{or} \quad x = -2$

(e) $3x^2 - 22x + 7 = 0$

This quadratic doesn't factor easily, so we'll use the quadratic formula again. Here, $a = 3$, $b = -22$, and $c = 7$. Plugging into the quadratic formula:

$x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4(3)(7)}}{2(3)} = \frac{22 \pm \sqrt{484 - 84}}{6} = \frac{22 \pm \sqrt{400}}{6}$

$x = \frac{22 \pm 20}{6}$

So, the solutions are:

$x = \frac{22 + 20}{6} = 7 \quad \text{or} \quad x = \frac{22 - 20}{6} = \frac{2}{6} = \frac{1}{3}$

2. Determine the roots of each equation.

(a) $y = x^2 + 9x + 3$

We'll use the quadratic formula again for this. Here, $a = 1$, $b = 9$, and $c = 3$:

$x = \frac{-9 \pm \sqrt{9^2 - 4(1)(3)}}{2(1)} = \frac{-9 \pm \sqrt{81 - 12}}{2} = \frac{-9 \pm \sqrt{69}}{2}$

The roots are:

$x = \frac{-9 \pm \sqrt{69}}{2}$

These are irrational solutions.

(b) $y = 3x^2 + 24x - 6$

We use the quadratic formula again. Here, $a = 3$, $b = 24$, and $c = -6$:

$x = \frac{-24 \pm \sqrt{24^2 - 4(3)(-6)}}{2(3)} = \frac{-24 \pm \sqrt{576 + 72}}{6} = \frac{-24 \pm \sqrt{648}}{6} = \frac{-24 \pm 18\sqrt{2}}{6}$

The solutions are:

$x = \frac{-24 + 18\sqrt{2}}{6} \quad \text{or} \quad x = \frac{-24 - 18\sqrt{2}}{6}$

If you need any further explanations or have questions, feel free to ask!

Here are 5 related questions you might find helpful:

1. How can you use the quadratic formula to solve any quadratic equation?
2. What are the conditions for factoring a quadratic equation easily?
3. How do irrational solutions affect the graph of a quadratic equation?
4. What are some real-life applications of solving quadratic equations?
5. How does completing the square compare to using the quadratic formula?

Tip: Always double-check your factorization by multiplying the factors back together to ensure you recover the original quadratic!