z^2=12z-27

Solution for z^2=12z-27 equation:

z^2=12z-27

We move all terms to the left:

z^2-(12z-27)=0

We get rid of parentheses

z^2-12z+27=0

a = 1; b = -12; c = +27;
Δ = b2-4ac
Δ = -122-4·1·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6}{2*1}=\frac{6}{2}
=3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6}{2*1}=\frac{18}{2}
=9
$