#### Answer

$x=\left\{ 1,6 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
(x+3)(x-6)=(2x+2)(x-6)
,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
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(-6)+3(x)+3(-6)=2x(x)+2x(-6)+2(x)+2(-6)
\\\\
x^2-6x+3x-18=2x^2-12x+2x-12
\\\\
(x^2-2x^2)+(-6x+3x+12x-2x)+(-18+12)=0
\\\\
-x^2+7x-6=0
\\\\
-1(-x^2+7x-6)=0(-1)
\\\\
x^2-7x+6=0
.\end{array}
To factor the trinomial expression above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ In the expression above, the value of $c$ is $
6
$ and the value of $b$ is $
-7
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,6 \}, \{ 2,3 \},
\\
\{ -1,-6 \}, \{ -2,-3 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-1,-6
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x-1)(x-6)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x-1=0
\text{ OR }
x-6=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x-1=0
\\\\
x=1
\\\\\text{ OR }\\\\
x-6=0
\\\\
x=6
.\end{array}
Hence, the solutions are $
x=\left\{ 1,6 \right\}
.$