We have to calculate the zeros of the function:
[tex]f(x)=x^2-4x+6[/tex]We apply the quadratic formula:
[tex]\begin{gathered} x=\frac{-(-4)}{2\cdot1}\pm\frac{\sqrt[]{(-4)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{4}{2}\pm\frac{\sqrt[]{16-24}}{2} \\ x=2\pm\sqrt[]{\frac{-8}{4}} \\ x=2\pm\sqrt[]{-2} \\ x=2\pm\sqrt[]{2\cdot(-1)} \\ x=2\pm\sqrt[]{2}\cdot i \\ x\approx2\pm1.414\cdot i \end{gathered}[/tex]The roots are imaginary, and have a value of x1=2+1.414i and x2=2-1.414i.
