Solve the given quadratic equation by completing the square. $x^2+3x-5=0$

• A. $x=-\(\frac\)32,x=\(\frac\)52$
• B. $x=-\(\frac\)32,x=\(\sqrt{29}\)$
• C. $x=\(\frac{-3+\sqrt{29}\)}{2},x=\(\frac{-3-\sqrt{29}\)}{2}$
• D. $x=\(\frac{3+\sqrt{29}\)}{2},x=\(\frac{3-\sqrt{29}\)}{2}$

Answer: (C)

Verified step by step guidance

1. Start with the quadratic equation: \( x^2 + 3x - 5 = 0 \). The goal is to complete the square to solve for \( x \).
2. Move the constant term to the other side of the equation: \( x^2 + 3x = 5 \).
3. To complete the square, take half of the coefficient of \( x \), which is \( \frac{3}{2} \), and square it: \( \left( \frac{3}{2} \right)^2 = \frac{9}{4} \).
4. Add \( \frac{9}{4} \) to both sides of the equation to form a perfect square trinomial on the left: \( x^2 + 3x + \frac{9}{4} = 5 + \frac{9}{4} \).
5. Rewrite the left side as a squared binomial: \( \left( x + \frac{3}{2} \right)^2 = \frac{29}{4} \). Now, solve for \( x \) by taking the square root of both sides and then isolating \( x \).