(z-4)(z-4)=64

Solution for (z-4)(z-4)=64 equation:

(z-4)(z-4)=64

We move all terms to the left:

(z-4)(z-4)-(64)=0

We multiply parentheses ..

(+z^2-4z-4z+16)-64=0

We get rid of parentheses

z^2-4z-4z+16-64=0

We add all the numbers together, and all the variables

z^2-8z-48=0

a = 1; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·1·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*1}=\frac{-8}{2}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*1}=\frac{24}{2}
=12
$