(3-2z)*(3-z+(3z*3z))=0

Solution for (3-2z)*(3-z+(3z*3z))=0 equation:

(3-2z)(3-z+(3z*3z))=0

We add all the numbers together, and all the variables

(-2z+3)(3-z+(+3z*3z))=0


We calculate terms in parentheses: +(-2z+3)(3-z+(+3z*3z)), so:

-2z+3)(3-z+(+3z*3z)

determiningTheFunctionDomain
-2z-z+(+3z*3z)+3)(3

We add all the numbers together, and all the variables

-3z+(+3z*3z)

We get rid of parentheses

-3z+3z*3z

Wy multiply elements

9z^2-3z

Back to the equation:

+(9z^2-3z)


We get rid of parentheses

9z^2-3z=0

a = 9; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·9·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*9}=\frac{0}{18}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*9}=\frac{6}{18}
=1/3
$