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

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

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

We multiply parentheses ..

(+6z^2+6z+8z+8)=0

We get rid of parentheses

6z^2+6z+8z+8=0

We add all the numbers together, and all the variables

6z^2+14z+8=0

a = 6; b = 14; c = +8;
Δ = b2-4ac
Δ = 142-4·6·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*6}=\frac{-16}{12}
=-1+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*6}=\frac{-12}{12}
=-1
$