We need to solve the quadratic equation:

\[
(2w + 11)w = 63
\]

Expanding the left-hand side:

\[
2w^2 + 11w = 63
\]

Now, rewrite it in standard quadratic form:

\[
2w^2 + 11w - 63 = 0
\]

This is a quadratic equation, and we can solve it using the quadratic formula:

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

For this equation, \(a = 2\), \(b = 11\), and \(c = -63\).

Now, substitute these values into the quadratic formula:

\[
w = \frac{-11 \pm \sqrt{11^2 - 4(2)(-63)}}{2(2)}
\]

Simplify inside the square root:

\[
w = \frac{-11 \pm \sqrt{121 + 504}}{4}
\]

\[
w = \frac{-11 \pm \sqrt{625}}{4}
\]

\[
w = \frac{-11 \pm 25}{4}
\]

Now, solve for both possible values of \(w\):

1. \(w_1 = \frac{-11 + 25}{4} = \frac{14}{4} = 3.5\)
2. \(w_2 = \frac{-11 - 25}{4} = \frac{-36}{4} = -9\)

So, the solutions are \(w = 3.5\) and \(w = -9\).

Would you like further details on solving quadratic equations?

Here are 5 related questions:
1. How do you factor quadratic equations?
2. What are the applications of the quadratic formula in real life?
3. How do you complete the square for a quadratic equation?
4. What happens when the discriminant is negative in the quadratic formula?
5. Can a quadratic equation have complex roots?

**Tip**: Always check the discriminant (\(b^2 - 4ac\)) to know how many real solutions exist!

Learn how to solve the quadratic equation (2w + 11)w = 63 step-by-step. This problem involves expanding and simplifying a quadratic equation, applying the quadratic formula, and solving for w. Ideal for students in Grades 8-10 studying algebra and quadratic equations.

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