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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-32x+32=0

a = 6; b = -32; c = +32;
Δ = b2-4ac
Δ = -322-4·6·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*6}=\frac{16}{12}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*6}=\frac{48}{12}
=4
$