Solve for x the quadratic equation `x^(2)-4x-8=0`. Give your answer correct to three significant figures.

Answer

To solve the quadratic equation \( x^2 - 4x - 8 = 0 \), we will use the quadratic formula, also known as the Sridharacharya formula. The quadratic formula is given by:

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

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).

### Step 1: Identify the coefficients
From the equation \( x^2 - 4x - 8 = 0 \), we can identify:
- \( a = 1 \)
- \( b = -4 \)
- \( c = -8 \)

### Step 2: Substitute the coefficients into the formula
Now we substitute \( a \), \( b \), and \( c \) into the quadratic formula:

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]

### Step 3: Simplify the expression
Calculating the components:

1. Calculate \( -(-4) \):
\[ -(-4) = 4 \]

2. Calculate \( (-4)^2 \):
\[ (-4)^2 = 16 \]

3. Calculate \( 4 \cdot 1 \cdot (-8) \):
\[ 4 \cdot 1 \cdot (-8) = -32 \]

4. Now, calculate \( b^2 - 4ac \):
\[ 16 - (-32) = 16 + 32 = 48 \]

5. Now, substitute back into the formula:
\[ x = \frac{4 \pm \sqrt{48}}{2} \]

### Step 4: Calculate the square root
Now we calculate \( \sqrt{48} \):
\[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \]

### Step 5: Substitute the square root back into the formula
Now substituting \( \sqrt{48} \) back into the equation:
\[ x = \frac{4 \pm 4\sqrt{3}}{2} \]

### Step 6: Simplify the expression
Divide each term by 2:
\[ x = 2 \pm 2\sqrt{3} \]

### Step 7: Calculate the two possible values for \( x \)
1. Calculate \( x_1 = 2 + 2\sqrt{3} \):
\[ 2 + 2\sqrt{3} \approx 2 + 2 \cdot 1.732 \approx 2 + 3.464 \approx 5.464 \]

2. Calculate \( x_2 = 2 - 2\sqrt{3} \):
\[ 2 - 2\sqrt{3} \approx 2 - 3.464 \approx -1.464 \]

### Step 8: Round to three significant figures
Thus, the two solutions rounded to three significant figures are:
- \( x_1 \approx 5.46 \)
- \( x_2 \approx -1.46 \)

### Final Answers
The solutions to the equation \( x^2 - 4x - 8 = 0 \) are:
\[ x \approx 5.46 \quad \text{and} \quad x \approx -1.46 \]