# How do you solve  x^2-5x=-4 by completing the square?

Apr 30, 2015

In this way:

${x}^{2} - 5 x = - 4 \Rightarrow {x}^{2} - 5 x + \frac{25}{4} - \frac{25}{4} = - 4 \Rightarrow$

${\left(x - \frac{5}{2}\right)}^{2} = \frac{25}{4} - 4 \Rightarrow {\left(x - \frac{5}{2}\right)}^{2} = \frac{25 - 16}{4} \Rightarrow$

${\left(x - \frac{5}{2}\right)}^{2} = \frac{9}{4} \Rightarrow x - \frac{5}{2} = \pm \frac{3}{2} \Rightarrow x = \frac{5}{2} \pm \frac{3}{2} \Rightarrow$

${x}_{1} = \frac{5}{2} - \frac{3}{2} = 1$

${x}_{2} = \frac{5}{2} + \frac{3}{2} = 4$.