(2x-5)(4x-8)=0

Solution for (2x-5)(4x-8)=0 equation:

(2x-5)(4x-8)=0

We multiply parentheses ..

(+8x^2-16x-20x+40)=0

We get rid of parentheses

8x^2-16x-20x+40=0

We add all the numbers together, and all the variables

8x^2-36x+40=0

a = 8; b = -36; c = +40;
Δ = b2-4ac
Δ = -362-4·8·40
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{-(-36)-4}{2*8}=\frac{32}{16}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*8}=\frac{40}{16}
=2+1/2
$