To solve the quadratic equation \( 14x^2 + 4x + 14 = -6x^2 \), let's follow these steps:

1. **Move all terms to one side of the equation**:  
   Add \( 6x^2 \) to both sides to combine the quadratic terms.
   \[
   14x^2 + 6x^2 + 4x + 14 = 0
   \]
   Simplifying, we get:
   \[
   20x^2 + 4x + 14 = 0
   \]

2. **Simplify the equation**:  
   Divide the entire equation by 2 to make it simpler:
   \[
   10x^2 + 2x + 7 = 0
   \]

3. **Apply the quadratic formula**:  
   The quadratic formula is given by:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 10 \), \( b = 2 \), and \( c = 7 \).

4. **Calculate the discriminant**:
   \[
   b^2 - 4ac = 2^2 - 4(10)(7) = 4 - 280 = -276
   \]
   The discriminant is negative, which means the equation has no real solutions, only complex solutions.

5. **Find the complex solutions**:
   \[
   x = \frac{-2 \pm \sqrt{-276}}{20}
   \]
   Simplify the square root:
   \[
   x = \frac{-2 \pm \sqrt{276}i}{20} 
   \]
   Further simplify:
   \[
   x = \frac{-2 \pm 2\sqrt{69}i}{20} = \frac{-1 \pm \sqrt{69}i}{10}
   \]

So, the solutions are:
\[
x = \frac{-1 + \sqrt{69}i}{10} \quad \text{and} \quad x = \frac{-1 - \sqrt{69}i}{10}
\]

Would you like more details on any of these steps, or do you have any other questions?

Here are five related questions:
1. How do you determine if a quadratic equation has real or complex solutions?
2. What are the implications of a negative discriminant in real-world problems?
3. How do you simplify complex numbers after applying the quadratic formula?
4. What does it mean when a quadratic equation has two complex conjugate solutions?
5. Can you graph a quadratic equation with complex solutions?

**Tip:** Always check the discriminant first when solving a quadratic equation, as it tells you whether the solutions are real or complex.

Learn how to solve the quadratic equation 14x^2 + 4x + 14 = -6x^2 with detailed step-by-step instructions. This problem involves complex numbers and the quadratic formula theorem. Suitable for high school students in Grades 10-12.

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