## Algebra 2 (1st Edition)

$x=8$ or $x=6$
Given: $y=x^2-14x+48$ The function is in quadratic form: $y=ax^2+bx+c$ with $a=1, b=-14, c=48$ Find x by using: $x=\frac{-b \pm \sqrt b^2-4ac}{2a}=\frac{-(-14)\pm\sqrt (-14)^2-4.1.48}{2.1}=\frac{14\pm2}{2}$ $x=\frac{14+2}{2}$ or $x=\frac{14-2}{2}$ $x=8$ or $x=6$