## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=\left\{ -3,1 \right\}$
Using the properties of equality, the given expression, $3x^2-2x=9-8x ,$ is equivalent to \begin{array}{l}\require{cancel} 3x^2-2x+8x-9=0 \\\\ 3x^2+6x-9=0 .\end{array} Factoring the above equation, $3x^2+6x-9=0 ,$ results to \begin{array}{l}\require{cancel} 3(x^2+2x-3)=0 \\\\ x^2+2x-3=0 \\\\ (x+3)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $(x+3)(x-1)=0 ,$ are \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 ,\\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $x=\left\{ -3,1 \right\} .$