(2x-5)(2x-3)/2=0

Solution for (2x-5)(2x-3)/2=0 equation:

(2x-5)(2x-3)/2=0

We multiply parentheses ..

(+4x^2-6x-10x+15)/2=0

We multiply all the terms by the denominator


(+4x^2-6x-10x+15)=0

We get rid of parentheses

4x^2-6x-10x+15=0

We add all the numbers together, and all the variables

4x^2-16x+15=0

a = 4; b = -16; c = +15;
Δ = b2-4ac
Δ = -162-4·4·15
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{-(-16)-4}{2*4}=\frac{12}{8}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*4}=\frac{20}{8}
=2+1/2
$