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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-20x+16=0

a = 6; b = -20; c = +16;
Δ = b2-4ac
Δ = -202-4·6·16
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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