The given function is
[tex]y=x^2-8x+12\text{ }[/tex]
To find the zeros, we use the quadratic formula.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where a = 1, b = -8, and c = 12.
[tex]\begin{gathered} x=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(1)(12)}_{}}{2(1)}=\frac{8\pm\sqrt[]{64-48}}{2} \\ x=\frac{8\pm\sqrt[]{16}}{2}=\frac{8\pm4}{2}=4\pm2 \\ x_1=4+2=6 \\ x_2=4-2=2 \end{gathered}[/tex]
Therefore, the zeros x = 6 and x = 2.
