Answer
$x=\left\{ \dfrac{-3-\sqrt{33}}{2},\dfrac{-3+\sqrt{33}}{2}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
x=\dfrac{2(x+3)}{x+5}
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x=\dfrac{2(x)+2(3)}{x+5}
\\\\
x=\dfrac{2x+6}{x+5}
.\end{array}
Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x(x+5)=1(2x+6)
\\\\
x^2+5x=2x+6
.\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2+5x-2x-6=0
\\\\
x^2+3x-6=0
.\end{array}
The quadratic equation above has $a=
1
, b=
3
, c=
-6
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-3\pm\sqrt{3^2-4(1)(-6)}}{2(1)}
\\\\
x=\dfrac{-3\pm\sqrt{9+24}}{2}
\\\\
x=\dfrac{-3\pm\sqrt{33}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-3-\sqrt{33}}{2}
\\\\\text{OR}\\\\
x=\dfrac{-3+\sqrt{33}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{-3-\sqrt{33}}{2},\dfrac{-3+\sqrt{33}}{2}
\right\}
.$
