32=12+4z(z-1)

Solution for 32=12+4z(z-1) equation:

32=12+4z(z-1)

We move all terms to the left:

32-(12+4z(z-1))=0


We calculate terms in parentheses: -(12+4z(z-1)), so:

12+4z(z-1)

determiningTheFunctionDomain
4z(z-1)+12

We multiply parentheses

4z^2-4z+12

Back to the equation:

-(4z^2-4z+12)


We get rid of parentheses

-4z^2+4z-12+32=0

We add all the numbers together, and all the variables

-4z^2+4z+20=0

a = -4; b = 4; c = +20;
Δ = b2-4ac
Δ = 42-4·(-4)·20
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{21}}{2*-4}=\frac{-4-4\sqrt{21}}{-8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{21}}{2*-4}=\frac{-4+4\sqrt{21}}{-8}
$