## College Algebra (10th Edition)

$x=\displaystyle \frac{3}{2}$ or $x=-2$
We solve: $(x-1)(2x+3)=3$ $2x^{2}+x-3=3$ $2x^{2}+x-3-3=0$ $2x^{2}+x-6=0$ $(2x-3)(x+2)=0$ $(2x-3)=0$ or $(x+2)=0$ $x=\displaystyle \frac{3}{2}$ or $x=-2$