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We are given the quadratic equation:

\[
6x^2 - 35 = -11x
\]

First, let's rearrange this equation into standard quadratic form \( ax^2 + bx + c = 0 \):

\[
6x^2 + 11x - 35 = 0
\]

Here, \( a = 6 \), \( b = 11 \), and \( c = -35 \).

Now, we'll apply the quadratic formula:

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

Substitute the values of \( a \), \( b \), and \( c \):

\[
x = \frac{-11 \pm \sqrt{11^2 - 4(6)(-35)}}{2(6)}
\]

Let's compute the discriminant first:

\[
b^2 - 4ac = 11^2 - 4(6)(-35) = 121 + 840 = 961
\]

So the quadratic formula becomes:

\[
x = \frac{-11 \pm \sqrt{961}}{12}
\]

Since \( \sqrt{961} = 31 \), we now have:

\[
x = \frac{-11 \pm 31}{12}
\]

This gives us two possible values for \( x \):

1. \( x_1 = \frac{-11 + 31}{12} = \frac{20}{12} = \frac{5}{3} \)
2. \( x_2 = \frac{-11 - 31}{12} = \frac{-42}{12} = -\frac{7}{2} \)

Thus, the two solutions for \( x \) are:

\[
x_1 = \frac{5}{3}, \quad x_2 = -\frac{7}{2}
\]

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How can you verify the solutions to a quadratic equation?
2. What is the discriminant, and how does it determine the nature of the roots?
3. How would the solutions change if the equation had no real solutions?
4. Can the quadratic formula be used for equations with complex numbers?
5. How does factoring relate to solving quadratic equations?

**Tip:** Always simplify the discriminant before proceeding with the quadratic formula—it makes the calculations smoother!

Using the quadratic formula, solve the equation below to find the two possible values of x: 6x^2 - 35 = -11x.

Solve the quadratic equation 6x^2 - 35 = -11x using the quadratic formula. This step-by-step solution involves rearranging the equation, calculating the discriminant, and finding the roots. Suitable for students in Grades 8-10.

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