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

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

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

We multiply parentheses ..

(+6x^2-10x-9x+15)=0

We get rid of parentheses

6x^2-10x-9x+15=0

We add all the numbers together, and all the variables

6x^2-19x+15=0

a = 6; b = -19; c = +15;
Δ = b2-4ac
Δ = -192-4·6·15
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*6}=\frac{18}{12}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*6}=\frac{20}{12}
=1+2/3
$