College Algebra 7th Edition

$x=4\pm\sqrt{15}$
We complete the square: $x^{2}-8x+1=0$ $x^{2}-8x=-1$ $x^{2}-8x+16=-1+16$ $(x-4)^{2}=15$ $x-4=\pm\sqrt{15}$ $x=4\pm\sqrt{15}$