Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decims places (if necessary). $$ y=x^{2}+10 x-96 $$

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To find the **roots** and the **vertex** of the quadratic equation $$ y = x^{2} + 10x - 96 $$, we'll use the following methods:

### 1. Finding the Roots:
The roots of a quadratic equation $$ ax^{2} + bx + c = 0 $$ can be found using the **Quadratic Formula**:
$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$
For the given equation:
- $$ a = 1 $$
- $$ b = 10 $$
- $$ c = -96 $$

Plugging these values into the formula:
$$
x = \frac{-10 \pm \sqrt{10^{2} - 4(1)(-96)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 384}}{2} = \frac{-10 \pm \sqrt{484}}{2}
$$
$$
\sqrt{484} = 22
$$
So,
$$
x = \frac{-10 \pm 22}{2}
$$
This gives two solutions:
$$
x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{-32}{2} = -16
$$

**Roots:** $$ x = 6 $$ and $$ x = -16 $$

### 2. Finding the Vertex:
The vertex $$(h, k)$$ of a quadratic equation $$ y = ax^{2} + bx + c $$ is given by:
$$
h = -\frac{b}{2a}
$$
$$
k = y(h) = a h^{2} + b h + c
$$

Calculating $$ h $$:
$$
h = -\frac{10}{2(1)} = -5
$$

Calculating $$ k $$:
$$
k = (-5)^{2} + 10(-5) - 96 = 25 - 50 - 96 = -121
$$

**Vertex:** $$ (-5.000, -121.000) $$

### **Summary:**
- **Roots:** $$ x = 6 $$ and $$ x = -16 $$
- **Vertex:** $$ (-5.000,\ -121.000) $$
**Roots:** $$ x = 6 $$ and $$ x = -16 $$ **Vertex:** $$ (-5.000, -121.000) $$

Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decims places (if necessary). \[ y=x^{2}+10 x-96 \]

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