$$ x^2-\sqrt2x-5=0\text{ The solution set is ​\textbraceleft enter your response here​\textbraceright. ​(Simplify your answer. Type an integer or decimal rounded to two decimal places as needed. Use a comma to separate answers as​ needed.)} $$

To solve the quadratic equation $ x^2 - \sqrt{2}x - 5 = 0 $ using the quadratic formula, we start with the standard form of the quadratic equation:

\[
ax^2 + bx + c = 0
\]

Here, $ a = 1 $, $ b = -\sqrt{2} $, and $ c = -5 $.

The quadratic formula is:

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

Substitute the values of $ a $, $ b $, and $ c $:

\[
x = \frac{-(-\sqrt{2}) \pm \sqrt{(-\sqrt{2})^2 - 4(1)(-5)}}{2(1)}
\]

Simplify this step by step:

1. Calculate the discriminant:

\[
(-\sqrt{2})^2 = 2
\]

\[
4ac = 4(1)(-5) = -20
\]

\[
b^2 - 4ac = 2 + 20 = 22
\]

2. Plug the discriminant back into the quadratic formula:

\[
x = \frac{\sqrt{2} \pm \sqrt{22}}{2}
\]

3. Simplify further:

\[
x_1 = \frac{\sqrt{2} + \sqrt{22}}{2}
\]

\[
x_2 = \frac{\sqrt{2} - \sqrt{22}}{2}
\]

4. Calculate the approximate decimal values (rounded to two decimal places):

\[
x_1 \approx \frac{1.41 + 4.69}{2} = \frac{6.10}{2} = 3.05
\]

\[
x_2 \approx \frac{1.41 - 4.69}{2} = \frac{-3.28}{2} = -1.64
\]

Thus, the solution set is:

\[
\{3.05, -1.64\}
\]

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

Here are 5 relative questions:

1. What is the discriminant and how does it affect the number of real solutions?
2. How does the quadratic formula apply to equations where the coefficient $ a $ is not 1?
3. Can the quadratic formula be used for complex solutions? If so, how?
4. What are some alternative methods to solve quadratic equations besides the quadratic formula?
5. How would the solution set change if $ c $ were positive instead of negative?

**Tip:** Always check the discriminant first before proceeding with the quadratic formula, as it will tell you whether real solutions exist.

Learn how to solve the quadratic equation x^2 - √2x - 5 = 0 with a detailed step-by-step solution. Understand the quadratic formula and its application to solve equations involving radicals and negative coefficients. Suitable for students in Grades 10-12.

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