The equation in the image is:

$\sqrt{-x^2 + 10x} = 2$

Steps to Solve:

1. Square Both Sides: Eliminate the square root by squaring both sides of the equation: $-x^2 + 10x = 2^2$ $-x^2 + 10x = 4$

2. Rearrange into Standard Form: Bring all terms to one side: $x^2 - 10x + 4 = 0$

3. Solve the Quadratic Equation: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -10$, $c = 4$. Substitute these values: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{10 \pm \sqrt{100 - 16}}{2}$ $x = \frac{10 \pm \sqrt{84}}{2}$ Simplify: $x = \frac{10 \pm 2\sqrt{21}}{2}$ $x = 5 \pm \sqrt{21}$

   The solutions are: $x_1 = 5 + \sqrt{21}, \quad x_2 = 5 - \sqrt{21}$

4. Check for Validity: Since the original equation contains a square root, we must check that both solutions satisfy the domain: $-x^2 + 10x \geq 0$ Factorize: $x(-x + 10) \geq 0$ Solve for intervals: $x \in [0, 10]$

   Check the solutions:

   • $x_1 = 5 + \sqrt{21} > 10$ (invalid as it exceeds the domain).
   • $x_2 = 5 - \sqrt{21} \in [0, 10]$ (valid).

Final Answer:

$x = 5 - \sqrt{21}$

Would you like a more detailed breakdown or have any questions? Here are some related questions you might find interesting:

1. How does squaring both sides of an equation affect the validity of solutions?
2. What is the significance of checking the domain for square root functions?
3. Can you generalize solving equations with square roots to other forms like $\sqrt{ax + b} = c$?
4. How can graphing help in verifying solutions to quadratic equations?
5. What are the properties of the discriminant in the quadratic formula?

Tip: Always check the domain of the original equation, especially when working with square roots or even powers.