Transpose -9 to the left to end up with the standard form of the quadratic equation: [tex]z^2-16z+9=0[/tex] Since it does not appear to have rational roots, we proceed with the quadratic formula, where a=1, b=-16, c=9 and substitute in the quadratic formula: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] [tex]=\frac{16\pm\sqrt{(-16)^2-4*1*9}}{2}[/tex] [tex]=\frac{16\pm\sqrt{256-36}}{2}[/tex] [tex]=8\pm\sqrt{55}[/tex] i.e. z=8+sqrt(55) or 8-sqrt(55)
