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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3z^2+14z-24=0

a = 3; b = 14; c = -24;
Δ = b2-4ac
Δ = 142-4·3·(-24)
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{-(14)-22}{2*3}=\frac{-36}{6}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+22}{2*3}=\frac{8}{6}
=1+1/3
$