(2x-5)(2x-1)=0

Solution for (2x-5)(2x-1)=0 equation:

(2x-5)(2x-1)=0

We multiply parentheses ..

(+4x^2-2x-10x+5)=0

We get rid of parentheses

4x^2-2x-10x+5=0

We add all the numbers together, and all the variables

4x^2-12x+5=0

a = 4; b = -12; c = +5;
Δ = b2-4ac
Δ = -122-4·4·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*4}=\frac{4}{8}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*4}=\frac{20}{8}
=2+1/2
$