#### Answer

$x=\left\{ -\dfrac{1}{2},4 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
2x^2=7x+4
,$ 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=7x+4
\\\\
2x^2-7x-4=0
\\\\
(2x+1)(x-4)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
2x+1=0
\text{ OR }
x-4=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
2x+1=0
\\\\
2x=-1
\\\\
x=-\dfrac{1}{2}
\\\\\text{ OR }\\\\
x-4=0
\\\\
x=4
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{1}{2},4 \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
2x^2-7x-4
,$ 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(-4)=-8
$ and the value of $b$ is $
-7
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-8\}, \{2,-4\},
\{-1,8\}, \{-2,4\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,-8
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x^2+x-8x-4
.\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+x)-(8x+4)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(2x+1)-4(2x+1)
.\end{array}
Factoring the $GCF=
(2x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2x+1)(x-4)
.\end{array}