The quadratic equation given is:

$x^2 + bx + c = 0$

It has solutions $-7 + 4i$ and $-7 - 4i$. These solutions are complex conjugates, meaning the sum and product of the roots will help us determine the coefficients $b$ and $c$.

Using Vieta's formulas:

For a quadratic equation $x^2 + bx + c = 0$ with roots $r_1$ and $r_2$:

1. The sum of the roots is $-b$.
2. The product of the roots is $c$.

In this case:

• The sum of the roots $(-7 + 4i) + (-7 - 4i) = -14$. So, $-b = -14$, which gives $b = 14$.

• The product of the roots $( -7 + 4i)( -7 - 4i) = (-7)^2 - (4i)^2 = 49 - (-16) = 49 + 16 = 65$. So, $c = 65$.

Therefore:

• $b = 14$
• $c = 65$

Would you like more details or clarifications?

Here are 5 related questions to explore this further:

1. What happens if the roots are not complex conjugates?
2. How do you derive the quadratic formula from Vieta's relations?
3. Can you use another method to find $b$ and $c$ from the given roots?
4. How are the coefficients related to the nature of the roots (real vs. complex)?
5. Can Vieta’s formulas be used for higher-degree polynomials?

Tip: For quadratic equations with complex conjugate roots, the imaginary parts always cancel out in the sum of the roots, leaving only the real part to determine $b$.