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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4z^2+6z+12z+18)=0

We get rid of parentheses

4z^2+6z+12z+18=0

We add all the numbers together, and all the variables

4z^2+18z+18=0

a = 4; b = 18; c = +18;
Δ = b2-4ac
Δ = 182-4·4·18
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*4}=\frac{-24}{8}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*4}=\frac{-12}{8}
=-1+1/2
$