how do I solve for x in x^2 - 8x + 12 = 0 using factoring?
\(x^2 - 8x + 12 = 0\\ x^2-2x-6x+12=0\\ x(x-2)-6(x-2)=0\\ (x-6)(x-2)=0\\ x=6 \vee x=2\)
\(\rm x^2-8x+12=0 \\ \\ x^2-2x-6x+12=0 \\ \\ x(x-2)-6(x-2)=0 \\ \\ (x-2)(x-6)=0 \\ \\ \bold{I:x-2=0 \longrightarrow x=2} \\ \bold{II:x-6=0 \longrightarrow x=6}\)
