## Algebra 1: Common Core (15th Edition)

$x=3$ or $x=-\frac{1}{2}$
$2x^2-5x-3=0$ $2x^2-6x+x-3=0$ $2x(x-3)+(x-3)=0$ $(x-3)(2x+1)=0$ $x=3$ or $x=-\frac{1}{2}$ The equation has two solutions, $x=3$ or $x=-\frac{1}{2}$