## College Algebra (10th Edition)

$y=2$ or $y=\frac{5}{3}$.
$3(1-y)^2+5(1-y)+2=0\\3(1+y^2-2y)+5-5y+2=0\\3+3y^2-6y+7-5y=0\\3y^2-11y+10=0\\3y^2-5y-6y+10=y(3y-5)-2(3y-5)=(y-2)(3y-5)=0$ Thus $y=2$ or $y=\frac{5}{3}$. Check: $3\cdot(-1)^2+5\cdot(-1)+2=3-5+2=0$, $3\cdot(-2/3)^2+5\cdot(-2/3)+2=4/3-10/3+2=0$