3(2z-4)10z=16

Solution for 3(2z-4)10z=16 equation:

3(2z-4)10z=16

We move all terms to the left:

3(2z-4)10z-(16)=0

We multiply parentheses

60z^2-120z-16=0

a = 60; b = -120; c = -16;
Δ = b2-4ac
Δ = -1202-4·60·(-16)
Δ = 18240
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:

$\sqrt{\Delta}=\sqrt{18240}=\sqrt{64*285}=\sqrt{64}*\sqrt{285}=8\sqrt{285}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{285}}{2*60}=\frac{120-8\sqrt{285}}{120}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{285}}{2*60}=\frac{120+8\sqrt{285}}{120}
$