Answer
$x=\left\{ -2-2i,-2+ 2i \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
(x+4)(x+2)=2x
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula to solve for $x.$
$\bf{\text{Solution Details:}}$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
x(x)+x(2)+4(x)+4(2)=2x
\\\\
x^2+2x+4x+8=2x
.\end{array}
In the form $ax^2+bx+c=0,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x^2+(2x+4x-2x)+8=0
\\\\
x^2+4x+8=0
.\end{array}
In the equation above, $a=
1
,$ $b=
4
,$ and $c=
8
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)}
\\\\
x=\dfrac{-4\pm\sqrt{16-32}}{2}
\\\\
x=\dfrac{-4\pm\sqrt{-16}}{2}
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and $i=\sqrt{-1},$ the equation above is equivalent to\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{16}}{2}
\\\\
x=\dfrac{-4\pm i\cdot\sqrt{(4)^2}}{2}
\\\\
x=\dfrac{-4\pm 4i}{2}
\\\\
x=\dfrac{\cancel2^{-2}\pm \cancel2^2i}{\cancel2^1}
\\\\
x=-2\pm 2i
.\end{array}
The solutions are $
x=\left\{ -2-2i,-2+ 2i \right\}
.$
