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

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

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

We multiply parentheses ..

(+8x^2-16x-12x+24)=0

We get rid of parentheses

8x^2-16x-12x+24=0

We add all the numbers together, and all the variables

8x^2-28x+24=0

a = 8; b = -28; c = +24;
Δ = b2-4ac
Δ = -282-4·8·24
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{-(-28)-4}{2*8}=\frac{24}{16}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*8}=\frac{32}{16}
=2
$