Question

Plz help and explain :)))

Answer 1

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Answer 2

Solve
$z^2-16z=-9$

Transpose -9 to the left to end up with the standard form of the quadratic equation:
$z^2-16z+9=0$
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:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$=\frac{16\pm\sqrt{(-16)^2-4*1*9}}{2}$
$=\frac{16\pm\sqrt{256-36}}{2}$
$=8\pm\sqrt{55}$
i.e.
z=8+sqrt(55) or 8-sqrt(55)