Answer
$x=\left\{ -\dfrac{5}{2},\dfrac{10}{3} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given eqution, $
6x^2=5x+50
,$ express the equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Next step is to equate each factor to zero (Zero Product Property). Finally, solve each equation.
$\bf{\text{Solution Details:}}$
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
6x^2-5x-50=0
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
6(-50)=-300
$ and the value of $b$ is $
-5
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
15,-20
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6x^2+15x-20x-50=0
.\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}
(6x^2+15x)-(20x+50)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3x(2x+5)-10(2x+5)=0
.\end{array}
Factoring the $GCF=
(2x+5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2x+5)(3x-10)=0
.\end{array}
Equating each factor to zero (Zero Product Property), the solutions to the equation above are
\begin{array}{l}\require{cancel}
2x+5=0
\\\\\text{OR}\\\\
3x-10=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2x+5=0
\\\\
2x=-5
\\\\
x=-\dfrac{5}{2}
\\\\\text{OR}\\\\
3x-10=0
\\\\
3x=10
\\\\
x=\dfrac{10}{3}
.\end{array}
Hence, $
x=\left\{ -\dfrac{5}{2},\dfrac{10}{3} \right\}
.$
