### Understanding the Quadratic Formula and Factoring

When dealing with quadratic equations, two fundamental methods you often encounter are the **quadratic formula** and **factoring**. Each method provides a way to find the solutions, or "roots", of a quadratic equation of the form:

$ax^2 + bx + c = 0$
#### Quadratic Formula

The quadratic formula is a straightforward way to solve any quadratic equation. The formula is:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$
Here's a breakdown of each component:

**$a$**, **$b$**, and **$c$** are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.
**$b^2 - 4ac$** is called the **discriminant**, and it determines the nature of the roots.
- If it's
**positive**, there are two distinct real roots.
- If it's
**zero**, there is exactly one real root (also called a repeated or double root).
- If it's
**negative**, there are two complex roots.

#### Factoring

Factoring involves rewriting the quadratic equation as a product of its linear factors, if they exist. Consider a quadratic equation in its simplified form:

$ax^2 + bx + c = 0$
To factor it, we assume it can be written as:

$(mx + n)(px + q) = 0$
Here, $m$, $n$, $p$, and $q$ are numbers that satisfy the condition $mp = a$, $nq = c$, and $mq + np = b$.

If you can identify such numbers, the equation factors into two binomials:

$(mx + n)(px + q) = 0$
Setting each factor to zero gives the solutions:

$mx + n = 0 \quad \text{or} \quad px + q = 0$
$x = -\frac{n}{m} \quad \text{and} \quad x = -\frac{q}{p}$
**Example:**

Consider the quadratic equation:

$2x^2 - 4x - 6 = 0$
**Using the Quadratic Formula:**
Here, $a = 2$, $b = -4$, and $c = -6$. Plugging these into the quadratic formula, we get:

$x = \frac{{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}}{2(2)}$
$x = \frac{{4 \pm \sqrt{16 + 48}}}{4}$
$x = \frac{{4 \pm \sqrt{64}}}{4}$
$x = \frac{{4 \pm 8}}{4}$
Thus, the solutions are:

$x = \frac{{4 + 8}}{4} = 3 \quad \text{and} \quad x = \frac{{4 - 8}}{4} = -1$
**Using Factoring:**
First, factor the equation:

$2x^2 - 4x - 6 = 0$
Divide through by 2:

$x^2 - 2x - 3 = 0$
Next, find two numbers that multiply to $-3$ and add to $-2$:

$(x - 3)(x + 1) = 0$
Setting each factor to zero:

$x - 3 = 0 \quad \text{or} \quad x + 1 = 0$
$x = 3 \quad \text{and} \quad x = -1$
**In summary:**
Both methods confirm the roots are $x = 3$ and $x = -1$. The **quadratic formula** works universally, while **factoring** relies on the quadratic being factorable into integers.