(6z+1)(-8z-5)=0

Solution for (6z+1)(-8z-5)=0 equation:

(6z+1)(-8z-5)=0

We multiply parentheses ..

(-48z^2-30z-8z-5)=0

We get rid of parentheses

-48z^2-30z-8z-5=0

We add all the numbers together, and all the variables

-48z^2-38z-5=0

a = -48; b = -38; c = -5;
Δ = b2-4ac
Δ = -382-4·(-48)·(-5)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-22}{2*-48}=\frac{16}{-96}
=-1/6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+22}{2*-48}=\frac{60}{-96}
=-5/8
$