## Algebra 2 (1st Edition)

$x= 4 \pm \sqrt{31}$
To complete the square, we add $(b/2)^2$ to both sides and subtract the constant. Adding $4^2 =16$ and adding 15, each to both sides. $x^2 - 8x +16 = 31$ $(x-4)^2 = 31$ $x= 4 \pm \sqrt{31}$