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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+6x^2-10x-9x+15)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-19x+15

Back to the equation:

-(6x^2-19x+15)


We get rid of parentheses

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