Solve for \( x \) by first completing the sque \( 0 x^{2}+5 x+4=0 \)

It looks like there's a small mistake in your equation—most likely a typo. The equation should be of the form \( ax^2 + bx + c = 0 \) where \( a \) is not zero. For instance, if you meant \( x^2 + 5x + 4 = 0 \), here's how you would complete the square. First, you rewrite the equation: \[ x^2 + 5x + 4 = 0 \] Next, isolate the \( x^2 \) and \( x \) terms: \[ x^2 + 5x = -4 \] Now, complete the square: Take half of the coefficient of \( x \) (which is \( 5 \)), square it (which gives \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \)), and add it to both sides: \[ x^2 + 5x + \frac{25}{4} = -4 + \frac{25}{4} \] This simplifies to: \[ (x + \frac{5}{2})^2 = \frac{9}{4} \] Taking the square root of both sides gives: \[ x + \frac{5}{2} = \pm \frac{3}{2} \] Finally, solve for \( x \): 1. \( x + \frac{5}{2} = \frac{3}{2} \) leads to \( x = \frac{3}{2} - \frac{5}{2} = -1 \) 2. \( x + \frac{5}{2} = -\frac{3}{2} \) leads to \( x = -\frac{3}{2} - \frac{5}{2} = -4 \) Thus, the solutions are \( x = -1 \) and \( x = -4 \).