Answer
The solutions are $x=-2$ and $x=1$.
Work Step by Step
RECALL:
$\log_a{b}=x \longrightarrow a^x=b $
where $b\gt0$
Use this rule in converting the equation to its equivalent exponential equation to have:
$2^1=x^2+x
\\2=x^2+x$
Subtract 2 to both sides to have:
$\\0=x^2+x-2
\\x^2+x-2=0$
Factor the trinomial to have:
$(x+2)(x-1)=0$
Equate each factor to zero then solve each equation to have:
$x+2=0 \text{ or } x-1=0
\\x=-2 \text{ or } x=1$
Note that in $\log_a{b}$, the value of $b$ has to be positive.
This means that in $\log_2{(x^2+x)}$, the value of $x^2+x$ must be positive.
Check if both solutions are valid to have:
For x = -2:
$x^2+x=(-2)^2+(-2) = 4-2 = 2$
For x=1:
$x^2+x=(1)^2+1=1+1=2$
Since both solutions make $x^2+x$ positive, then both are valid.
Therefore, the solutions to the given equation are -2 and 1.
