Solve
[tex]z^2-16z=-9[/tex]
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)
