## College Algebra (11th Edition)

$x=\left\{ -3,\dfrac{5}{2} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $2x^2+x-15=0 ,$ express the left side in factored form and equate each to factor to zero (Zero Product Property). Then use the properties of equality to isolate the variable in each equation. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (2x-5)(x+3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 2x-5=0 \\\\\text{OR}\\\\ x+3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x-5=0 \\\\ 2x=5 \\\\ x=\dfrac{5}{2} \\\\\text{OR}\\\\ x+3=0 \\\\ x=-3 .\end{array} Hence, the solutions are $x=\left\{ -3,\dfrac{5}{2} \right\} .$