z^2-16z-81=0

Solution for z^2-16z-81=0 equation:

z^2-16z-81=0

a = 1; b = -16; c = -81;
Δ = b2-4ac
Δ = -162-4·1·(-81)
Δ = 580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{145}}{2*1}=\frac{16-2\sqrt{145}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{145}}{2*1}=\frac{16+2\sqrt{145}}{2}
$