@aks.shan6 is it by chance C not sure though ?

**Note:** A better way to ask the question will be to attach you rough work (no matter how wrong you think it is) and then ask for suggestion - This way you’ll learn a lot more !

```
Subtract 28 from both sides:
```

x^2 + 8x + 12 - 28 <= 0

This simplifies to:

x^2 + 8x - 16 <= 0

```
To find the critical points, we set the expression equal to zero:
```

x^2 + 8x - 16 = 0

```
This is a quadratic equation that can be solved by factoring,
completing the square, or the quadratic formula.
We will use the quadratic formula:
```

x = \frac{[-b ± \sqrt{(b^2 - 4ac)}] }{ 2a}

In this case, a = 1, b = 8, and c = -16.

x = [-8 ± √(8^2 - 4(1)(-16))] / 2(1)

x = [-8 ± √(64 + 64)] / 2

x = [-8 ± √128] / 2

x = [-8 ± 8√2] / 2

```
Simplify:
```

x = -4 ± 4√2

So, the critical points are -4 + 4√2 and -4 - 4√2 .

-4√2 - 4 ≤ x ≤ 4√2 - 4

**Quantity A** :

4√2 - 4 +(-4√2) - 4 = -4 - 4 = -8

**Quantity B**

= -8

Why are we not using the other two methods?

Laziness and the quadratic formula is foolproof (you just have to substitute the values). Hence, very little work for me if I just use quadratic formula but you can go whichever way you find handy!

Thank you so much for your kind response!!