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

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

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

We multiply parentheses ..

(+6x^2+8x-12x-16)=0

We get rid of parentheses

6x^2+8x-12x-16=0

We add all the numbers together, and all the variables

6x^2-4x-16=0

a = 6; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·6·(-16)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*6}=\frac{24}{12}
=2
$