Algebra 2 (1st Edition)

$b^{2}-4ac=0$
$a=1,b=-6,c=9$ $b^{2}-4ac\qquad$ ...substitute $1$ for $a,\ -6$ for $b$ and $9$ for $c$. $=(-6)^{2}-4\cdot 1\cdot 9\qquad$ ...evaluate $(-6)^{2}$ $=36-4\cdot 1\cdot 9\qquad$ ...multiply before subtracting $=36-36$ $=0$