Solve the quadratic equation. Check the answers. \( \begin{array}{l}x^{2}+2 x-33=2 \\ \text { A. The solution(s) is/are } x= \\ \text { (Simplify your answer. Type an integer or a frection. Use no real solutions. }\end{array} \) Bhemma to separate answers as needed.)

First, let's rewrite the equation by moving all terms to one side: \( x^2 + 2x - 33 - 2 = 0 \) This simplifies to: \( x^2 + 2x - 35 = 0 \) Next, we can factor the quadratic equation: Look for two numbers that multiply to -35 and add to 2. These numbers are 7 and -5. So, we can factor the equation as: \( (x + 7)(x - 5) = 0 \) Setting each factor to zero gives us the solutions: \( x + 7 = 0 \quad \Rightarrow \quad x = -7 \) \( x - 5 = 0 \quad \Rightarrow \quad x = 5 \) The solutions are: **A. The solution(s) is/are** \( x = -7 \text{ and } x = 5 \) To check the answers, substitute \( x = -7 \) and \( x = 5 \) back into the original rearranged equation: For \( x = -7 \): \( (-7)^2 + 2(-7) - 35 = 49 - 14 - 35 = 0 \) (Correct) For \( x = 5 \): \( (5)^2 + 2(5) - 35 = 25 + 10 - 35 = 0 \) (Correct) Both solutions are verified!