Answer
$x=\left\{ -\dfrac{3}{2},1 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Express the given equation, $
2x^2=3-x
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
2x^2+x-3=0
\\\\
(x-1)(2x+3)
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
x-1=0
\text{ OR }
2x+3=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
x-1=0
\\\\
x=1
\\\\\text{ OR }\\\\
2x+3=0
\\\\
2x=-3
\\\\
x=-\dfrac{3}{2}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{3}{2},1 \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
2x^2+x-3
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
In the expression above the value of $ac$ is $
2(-3)=-6
$ and the value of $b$ is $
1
.$
The possible pairs of integers whose product is $ac$ 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 $\{
-2,3
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x^2-2x+3x-3
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(2x^2-2x)+(3x-3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2x(x-1)+3(x-1)
.\end{array}
Factoring the $GCF=
(x-1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x-1)(2x+3)
.\end{array}
