Use Newton's method to estimate the solutions of the equation $$ 7 x^{2}+x-1=0 $$. Start with $$ x_{0}=-1 $$ for the left solution and $$ x_{0}=1 $$ for the right solution. Find $$ x_{2} $$ in each case. Using Newton's method with $$ x_{0}=-1 $$, the third approximation, $$ x_{2} $$, to the left solution to $$ 7 x^{2}+x-1=0 $$ is (Round to four decimal places as needed.) Using Newton's method with $$ x_{0}=1 $$, the the third approximation, $$ x_{2} $$, to the right solution to $$ 7 x^{2}+x-1=0 $$ is (Round to four decimal places as needed.)

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To apply Newton's method to the equation $$ 7x^2 + x - 1 = 0 $$, we first need to define the function and its derivative.

1. **Define the function and its derivative:**
   - Let $$ f(x) = 7x^2 + x - 1 $$.
   - The derivative $$ f'(x) $$ is calculated as follows:
     $$
     f'(x) = \frac{d}{dx}(7x^2 + x - 1) = 14x + 1
     $$

2. **Newton's method formula:**
   The formula for Newton's method is given by:
   $$
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   $$

3. **Calculate $$ x_1 $$ and $$ x_2 $$ for both initial guesses:**

### For $$ x_0 = -1 $$:
- **Step 1: Calculate $$ x_1 $$**
  $$
  f(-1) = 7(-1)^2 + (-1) - 1 = 7 - 1 - 1 = 5
  $$
  $$
  f'(-1) = 14(-1) + 1 = -14 + 1 = -13
  $$
  $$
  x_1 = -1 - \frac{5}{-13} = -1 + \frac{5}{13} = -1 + 0.3846 \approx -0.6154
  $$

- **Step 2: Calculate $$ x_2 $$**
  $$
  f(-0.6154) = 7(-0.6154)^2 + (-0.6154) - 1
  $$
  First, calculate $$ (-0.6154)^2 $$:
  $$
  (-0.6154)^2 \approx 0.3795
  $$
  Now substitute:
  $$
  f(-0.6154) \approx 7(0.3795) - 0.6154 - 1 \approx 2.6565 - 0.6154 - 1 \approx 1.0411
  $$
  Now calculate $$ f'(-0.6154) $$:
  $$
  f'(-0.6154) = 14(-0.6154) + 1 \approx -8.6156 + 1 \approx -7.6156
  $$
  Now substitute into the Newton's method formula:
  $$
  x_2 = -0.6154 - \frac{1.0411}{-7.6156} \approx -0.6154 + 0.1364 \approx -0.4790
  $$

### For $$ x_0 = 1 $$:
- **Step 1: Calculate $$ x_1 $$**
  $$
  f(1) = 7(1)^2 + (1) - 1 = 7 + 1 - 1 = 7
  $$
  $$
  f'(1) = 14(1) + 1 = 14 + 1 = 15
  $$
  $$
  x_1 = 1 - \frac{7}{15} = 1 - 0.4667 \approx 0.5333
  $$

- **Step 2: Calculate $$ x_2 $$**
  $$
  f(0.5333) = 7(0.5333)^2 + (0.5333) - 1
  $$
  First, calculate $$ (0.5333)^2 $$:
  $$
  (0.5333)^2 \approx 0.2844
  $$
  Now substitute:
  $$
  f(0.5333) \approx 7(0.2844) + 0.5333 - 1 \approx 1.9908 + 0.5333 - 1 \approx 1.5241
  $$
  Now calculate $$ f'(0.5333) $$:
  $$
  f'(0.5333) = 14(0.5333) + 1 \approx 7.4662 + 1 \approx 8.4662
  $$
  Now substitute into the Newton's method formula:
  $$
  x_2 = 0.5333 - \frac{1.5241}{8.4662} \approx 0.5333 - 0.1805 \approx 0.3528
  $$

### Final Results:
- The third approximation $$ x_2 $$ for the left solution (starting with $$ x_0 = -1 $$) is approximately $$ -0.4790 $$.
- The third approximation $$ x_2 $$ for the right solution (starting with $$ x_0 = 1 $$) is approximately $$ 0.3528 $$.

Rounding to four decimal places:
- Left solution: $$ x_2 \approx -0.4790 $$
- Right solution: $$ x_2 \approx 0.3528 $$
For the left solution starting with $$ x_0 = -1 $$, the third approximation $$ x_2 $$ is approximately **-0.4790**. For the right solution starting with $$ x_0 = 1 $$, the third approximation $$ x_2 $$ is approximately **0.3528**.

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