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

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

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

We multiply parentheses ..

(+6x^2+4x+12x+8)=0

We get rid of parentheses

6x^2+4x+12x+8=0

We add all the numbers together, and all the variables

6x^2+16x+8=0

a = 6; b = 16; c = +8;
Δ = b2-4ac
Δ = 162-4·6·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*6}=\frac{-24}{12}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*6}=\frac{-8}{12}
=-2/3
$