z(z*2+1)=9+z*3

Solution for z(z*2+1)=9+z*3 equation:

z(z*2+1)=9+z*3

We move all terms to the left:

z(z*2+1)-(9+z*3)=0

We add all the numbers together, and all the variables

z(z*2+1)-(z*3+9)=0

We multiply parentheses

2z^2+z-(z*3+9)=0

We get rid of parentheses

2z^2+z-z*3-9=0

Wy multiply elements

2z^2+z-3z-9=0

We add all the numbers together, and all the variables

2z^2-2z-9=0

a = 2; b = -2; c = -9;
Δ = b2-4ac
Δ = -22-4·2·(-9)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{19}}{2*2}=\frac{2-2\sqrt{19}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{19}}{2*2}=\frac{2+2\sqrt{19}}{4}
$