2z(3z+8)=32

Solution for 2z(3z+8)=32 equation:

2z(3z+8)=32

We move all terms to the left:

2z(3z+8)-(32)=0

We multiply parentheses

6z^2+16z-32=0

a = 6; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·6·(-32)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-32}{2*6}=\frac{-48}{12}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+32}{2*6}=\frac{16}{12}
=1+1/3
$