(-6z+1)(4z-3)=0

Solution for (-6z+1)(4z-3)=0 equation:

(-6z+1)(4z-3)=0

We multiply parentheses ..

(-24z^2+18z+4z-3)=0

We get rid of parentheses

-24z^2+18z+4z-3=0

We add all the numbers together, and all the variables

-24z^2+22z-3=0

a = -24; b = 22; c = -3;
Δ = b2-4ac
Δ = 222-4·(-24)·(-3)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*-24}=\frac{-36}{-48}
=3/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*-24}=\frac{-8}{-48}
=1/6
$