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The equation provided is \( w^2 + 12w - 40 = 0 \). We are asked to find the solution from the provided options.

To solve this quadratic equation, we can either factor it, complete the square, or use the quadratic formula. Since factoring doesn't seem obvious here, we'll apply the **quadratic formula**, which is given by:

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

For the equation \( w^2 + 12w - 40 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = 12 \)
- \( c = -40 \)

Now, we substitute these values into the quadratic formula:

\[
w = \frac{-12 \pm \sqrt{12^2 - 4(1)(-40)}}{2(1)}
\]
\[
w = \frac{-12 \pm \sqrt{144 + 160}}{2}
\]
\[
w = \frac{-12 \pm \sqrt{304}}{2}
\]
\[
w = \frac{-12 \pm \sqrt{16 \times 19}}{2}
\]
\[
w = \frac{-12 \pm 4\sqrt{19}}{2}
\]
\[
w = -6 \pm 2\sqrt{19}
\]

Thus, the two possible solutions are:

\[
w = -6 + 2\sqrt{19} \quad \text{or} \quad w = -6 - 2\sqrt{19}
\]

From the options provided, **Option D**: \( -6 + 2\sqrt{19} \) is the correct solution.

Would you like more details on this solution, or do you have any further questions?

Here are five related questions for further practice:
1. Can you solve \( x^2 + 6x - 16 = 0 \) using the quadratic formula?
2. How do you derive the discriminant in the quadratic formula, and what does it tell you about the nature of the solutions?
3. How would you complete the square for \( w^2 + 12w - 40 = 0 \)?
4. Can you solve the quadratic equation by factoring instead of using the quadratic formula?
5. What are the possible solutions if the discriminant is zero in a quadratic equation?

**Tip:** The discriminant \( b^2 - 4ac \) determines whether the solutions are real, imaginary, or repeated!

Which of the following is a solution to the given equation: w^2 + 12w - 40 = 0?

Find out how to solve the quadratic equation w^2 + 12w - 40 = 0 using the quadratic formula. The solution involves finding the roots with detailed step-by-step instructions, identifying key concepts in algebra, radical expressions, and quadratic equations. Suitable for students in Grades 9-11.

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