(2x-8)(3x+4)=0

Solution for (2x-8)(3x+4)=0 equation:

(2x-8)(3x+4)=0

We multiply parentheses ..

(+6x^2+8x-24x-32)=0

We get rid of parentheses

6x^2+8x-24x-32=0

We add all the numbers together, and all the variables

6x^2-16x-32=0

a = 6; b = -16; c = -32;
Δ = b2-4ac
Δ = -162-4·6·(-32)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*6}=\frac{-16}{12}
=-1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*6}=\frac{48}{12}
=4
$