2z(3z+10)=16

Solution for 2z(3z+10)=16 equation:

2z(3z+10)=16

We move all terms to the left:

2z(3z+10)-(16)=0

We multiply parentheses

6z^2+20z-16=0

a = 6; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·6·(-16)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*6}=\frac{-48}{12}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*6}=\frac{8}{12}
=2/3
$