Answer
$x=\left\{ -3,4 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Express the given equation, $
2x^2-12-4x=x^2-3x
,$ 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 properties of equality and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(2x^2-x^2)+(-4x+3x)-12=0
\\\\
x^2-x-12=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 $
-12
$ and the value of $b$ is $
-1
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,-12 \}, \{ 2,-6 \}, \{ 3,-4 \},
\\
\{ -1,12 \}, \{ -2,6 \}, \{ -3,4 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
3,-4
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x+3)(x-4)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x+3=0
\text{ OR }
x-4=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x+3=0
\\\\
x=-3
\\\\\text{ OR }\\\\
x-4=0
\\\\
x=4
.\end{array}
Hence, the solutions are $
x=\left\{ -3,4 \right\}
.$
