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