### Explanation
The equation [tex]\((x-4)^2 - 28 = 8\)[/tex] has two solutions because it is a quadratic equation. Quadratic equations generally yield two solutions because they represent parabolas, which can intersect the x-axis at two points. To solve for [tex]\(x\)[/tex], we will isolate the variable by following a series of algebraic steps.
### Step-by-Step Solution
#### Step 1: Isolate the squared term
Start by isolating the term containing the square on one side of the equation.
[tex]\[
(x-4)^2 - 28 = 8
\][/tex]
To isolate [tex]\((x-4)^2\)[/tex], add 28 to both sides of the equation:
[tex]\[
(x-4)^2 - 28 + 28 = 8 + 28
\][/tex]
[tex]\[
(x-4)^2 = 36
\][/tex]
#### Step 2: Take the square root of both sides
To solve for [tex]\(x-4\)[/tex], take the square root of both sides of the equation. Remember that taking the square root of both sides yields both positive and negative values.
[tex]\[
\sqrt{(x-4)^2} = \pm \sqrt{36}
\][/tex]
[tex]\[
x-4 = \pm 6
\][/tex]
This gives us two cases to consider:
1. [tex]\(x - 4 = 6\)[/tex]
2. [tex]\(x - 4 = -6\)[/tex]
#### Step 3: Solve each case for [tex]\(x\)[/tex]
Case 1: [tex]\(x - 4 = 6\)[/tex]
Add 4 to both sides:
[tex]\[
x = 6 + 4
\][/tex]
[tex]\[
x = 10
\][/tex]
Case 2: [tex]\(x - 4 = -6\)[/tex]
Add 4 to both sides:
[tex]\[
x = -6 + 4
\][/tex]
[tex]\[
x = -2
\][/tex]
### Final Solutions
The solutions to the equation [tex]\((x-4)^2 - 28 = 8\)[/tex] are:
[tex]\[
x = 10 \quad \text{and} \quad x = -2
\][/tex]
These two solutions correspond to the points where the parabola defined by the left-hand side of the equation intersects the horizontal line defined by the right-hand side of the equation.
