-9.3z-z(2+1/3)+(-2/3)=z

Solution for -9.3z-z(2+1/3)+(-2/3)=z equation:

-9.3z-z(2+1/3)+(-2/3)=z

We move all terms to the left:

-9.3z-z(2+1/3)+(-2/3)-(z)=0

We add all the numbers together, and all the variables

-9.3z-z(1/3+2)-z+(-2/3)=0

We add all the numbers together, and all the variables

-10.3z-z(1/3+2)+(-2/3)=0

We multiply parentheses

-z^2-10.3z-2z+(-2/3)=0

We get rid of parentheses

-z^2-10.3z-2z-2/3=0

We multiply all the terms by the denominator


-z^2*3-(10.3z)*3-2z*3-2=0

We add all the numbers together, and all the variables

-z^2*3-(+10.3z)*3-2z*3-2=0

We multiply parentheses

-z^2*3-30z-2z*3-2=0

Wy multiply elements

-3z^2-30z-6z-2=0

We add all the numbers together, and all the variables

-3z^2-36z-2=0

a = -3; b = -36; c = -2;
Δ = b2-4ac
Δ = -362-4·(-3)·(-2)
Δ = 1272
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:

$\sqrt{\Delta}=\sqrt{1272}=\sqrt{4*318}=\sqrt{4}*\sqrt{318}=2\sqrt{318}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{318}}{2*-3}=\frac{36-2\sqrt{318}}{-6}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{318}}{2*-3}=\frac{36+2\sqrt{318}}{-6}
$