College Algebra 7th Edition

$3\pm 2\sqrt{2}$
We solve using the quadratic formula: $x^{2}-6x + 1 = 0$ $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^{2}-4(1)(1)}}{2(1)}=\frac{6\pm\sqrt{36-4}}{2}=\frac{6\pm\sqrt{32}}{2}=\frac{6\pm\sqrt{16*2}}{2}=\frac{6\pm 4\sqrt{2}}{2}=3\pm 2\sqrt{2}$