(2z-8)(3z+28)=0

Solution for (2z-8)(3z+28)=0 equation:

(2z-8)(3z+28)=0

We multiply parentheses ..

(+6z^2+56z-24z-224)=0

We get rid of parentheses

6z^2+56z-24z-224=0

We add all the numbers together, and all the variables

6z^2+32z-224=0

a = 6; b = 32; c = -224;
Δ = b2-4ac
Δ = 322-4·6·(-224)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-80}{2*6}=\frac{-112}{12}
=-9+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+80}{2*6}=\frac{48}{12}
=4
$