#### Answer

$x_{1}= 0$ and $x_{2}= -5$

#### Work Step by Step

Given $x^2+5x=0$
$a=1, \ b=5, \ c=0$
Using the quadratic formula: $\dfrac{-\ b \pm \sqrt{b^2-4ac}}{2a} ,$ we have:
$\dfrac{- 5 \pm \sqrt{5^2-4\times 1\times 0}}{2\times 1} = \dfrac{- 5 \pm \sqrt{25-0}}{2} = \dfrac{- 5 \pm \sqrt{25}}{2} = \dfrac{- 5 \pm 5}{2} $
Therefore, the solutions are $x_{1}= \dfrac{- 5 + 5}{2} = \dfrac{0}{2} = 0$ and $x_{2}= \dfrac{- 5 - 5}{2} = \dfrac{-10}{2} = -5$