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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3z^2+2z+12z+8)=0

We get rid of parentheses

3z^2+2z+12z+8=0

We add all the numbers together, and all the variables

3z^2+14z+8=0

a = 3; b = 14; c = +8;
Δ = b2-4ac
Δ = 142-4·3·8
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*3}=\frac{-24}{6}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*3}=\frac{-4}{6}
=-2/3
$