-66+z^2-19z=0

Solution for -66+z^2-19z=0 equation:

-66+z^2-19z=0

a = 1; b = -19; c = -66;
Δ = b2-4ac
Δ = -192-4·1·(-66)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-25}{2*1}=\frac{-6}{2}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+25}{2*1}=\frac{44}{2}
=22
$