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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4z^2+12z+2z+6)=0

We get rid of parentheses

4z^2+12z+2z+6=0

We add all the numbers together, and all the variables

4z^2+14z+6=0

a = 4; b = 14; c = +6;
Δ = b2-4ac
Δ = 142-4·4·6
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*4}=\frac{-24}{8}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*4}=\frac{-4}{8}
=-1/2
$