Answer
$x=-1\pm\sqrt{6}$
Work Step by Step
$x^{2}+2x-5=0$
Since we need to solve this by completing the square, let's take the independent term to the right side of the equation. If we do that, the equation becomes:
$x^{2}+2x=5$
We remember, that in order to complete this square, we add $(\dfrac{b}{2})^{2}$ to both sides of the equation. Also remember, that $b$ is always the coefficient of the first degree term. For this equation, $b=2$.
$x^{2}+2x+(\dfrac{2}{2})^{2}=5+(\dfrac{2}{2})^{2}$
$x^{2}+2x+1=6$
On the left side of the equation, we are left with a perfect square trinomial. So we factor it and the equation becomes:
$(x+1)^{2}=6$
Take the square root of both sides:
$\sqrt{(x+1)^{2}}=\sqrt{6}$
$x+1=\pm\sqrt{6}$
Solve for $x$:
$x=-1\pm\sqrt{6}$
