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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-1x-15

Back to the equation:

-(6x^2-1x-15)


We get rid of parentheses

-6x^2+1x+15=0

We add all the numbers together, and all the variables

-6x^2+x+15=0

a = -6; b = 1; c = +15;
Δ = b2-4ac
Δ = 12-4·(-6)·15
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*-6}=\frac{-20}{-12}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*-6}=\frac{18}{-12}
=-1+1/2
$