Answer
$x=-1+\sqrt{10}$ and $x= -1-\sqrt {10}$
Work Step by Step
$x^{2}+2x=9\implies x^{2}+2x-9=0$
Comparing $x^{2}+2x-9=0$ with $ax^{2}+bx+c=0$, we get
$a=1$, $b=2$ and $c=-9$.
The quadratic formula is
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Substituting the values, we get
$x=\frac{-(2) \pm \sqrt {(2)^{2}-4(1)(-9)}}{(2)(1)}=\frac{-2\pm \sqrt {40}}{2}=-1\pm \sqrt {10}$
The solutions of the equation are $x=-1+\sqrt{10}$ and $x= -1-\sqrt {10}$
