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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8x^2-20x+12=0

a = 8; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·8·12
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*8}=\frac{16}{16}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*8}=\frac{24}{16}
=1+1/2
$