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

$x=-$\frac$32,x=$\frac$52$

$x=-$\frac$32,x=$\sqrt{29}$$

$x=$\frac{-3+\sqrt{29}$}{2},x=$\frac{-3-\sqrt{29}$}{2}$

$x=$\frac{3+\sqrt{29}$}{2},x=$\frac{3-\sqrt{29}$}{2}$

Verified step by step guidance

Start with the quadratic equation: $ x^2 + 3x - 5 = 0 $. The goal is to complete the square to solve for $ x $.

Move the constant term to the other side of the equation: $ x^2 + 3x = 5 $.

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} $.

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} $.

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 $.

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