Answer
The solutions are $x=-1$ and $x=-7.5$.
Work Step by Step
Multiply $(x+6)^2$ to both sides of the equation to obtain:
$(x+6)^2\left(2-\dfrac{7}{x+6}\right)=\dfrac{15}{(x+6)^2} \cdot (x+6)^2
\\2(x+6)^2-(x+6)(7)=15
\\2(x+6)^2-7(x+6)-15=0$
Let
$u=x+6$
Rewrite the equation above using the variable $u$ to obtain:
$2u^2-7u-15=0$
Factor the trinomial to obtain:
$(2u+3)(u-5)=0$
Use the Zero-Factor Property by equating each unique factor to zero. then solve each equation:
\begin{array}{ccc}
&2u+3 = 0 &\text{ or } &u-5=0
\\&2u=-3 &\text{ or } &u=5
\\&u=-\frac{3}{2} &\text{ or } &u=5
\end{array}
Since $u=x+6$, then
\begin{array}{ccc}
&u= -\frac{3}{2} &\text{or} &u=5
\\&x+6=-\frac{3}{2} &\text{or} &x+6=5
\\&x=-\frac{3}{2}]-6 &\text{or} &x=5-6
\\&x=-7.5 &\text{or} &x=-1
\end{array}
Thus, the solutions are $x=-1$ and $x=-7.5$.
