Answer
$x=-2$ and $x=1$
Work Step by Step
$\log_{2}(x^{2}+x)=1$
Rewrite in exponential form:
$2^{1}=x^{2}+x$
$x^{2}+x=2$
Take the $2$ to the left side of the equation:
$x^{2}+x-2=0$
Solve this equation by factoring:
$(x+2)(x-1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+2=0$
$x=-2$
$x-1=0$
$x=1$
The initial equation is not undefined for either of these solutions, so our answer is:
$x=-2$ and $x=1$
