## Algebra and Trigonometry 10th Edition

Solutions: $x=1+\frac{\sqrt 2}{2}$ $x=1-\frac{\sqrt 2}{2}$
$6x^2-12x=-3$ (divide both sides by 6) $x^2-2x=-\frac{3}{6}$ $x^2-2(1)x+1^2=-\frac{1}{2}+1^2$ $(x-1)^2=\frac{1}{2}$ $x-1=±\sqrt {\frac{1}{2}}=±\frac{1}{\sqrt 2}\frac{\sqrt 2}{\sqrt 2}=±\frac{\sqrt 2}{2}$ $x=1+\frac{\sqrt 2}{2}$ or $x=1-\frac{\sqrt 2}{2}$