Okay, here we have this:
Considering the provided function, we are going to calculate the requested roots, so we obtain the following:
[tex]\begin{gathered} x^2-12x+32=0 \\ x_{1,2}=\frac{-(-12)\pm\sqrt{(-12^2)-4(1)(32)}}{2(1)} \\ x_{1,2}=\frac{12\pm\sqrt{144-128}}{2} \\ x_{1,2}=\frac{12\pm\sqrt{16}}{2} \\ x_{1,2}=\frac{12\pm4}{2} \\ x_{1,2}=6\pm2 \\ x_1=6+2,x_2=6-2 \\ x_1=8,x_2=4 \end{gathered}[/tex]Finally we obtain that the roots of the function are x=8 and x=4.
