2(5z-4)(z+3)=0

Solution for 2(5z-4)(z+3)=0 equation:

2(5z-4)(z+3)=0

We multiply parentheses ..

2(+5z^2+15z-4z-12)=0

We multiply parentheses

10z^2+30z-8z-24=0

We add all the numbers together, and all the variables

10z^2+22z-24=0

a = 10; b = 22; c = -24;
Δ = b2-4ac
Δ = 222-4·10·(-24)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-38}{2*10}=\frac{-60}{20}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*10}=\frac{16}{20}
=4/5
$