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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+6x^2+15x-4x-10)), so:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+11x-10

Back to the equation:

-(6x^2+11x-10)


We get rid of parentheses

-6x^2-11x+10=0

a = -6; b = -11; c = +10;
Δ = b2-4ac
Δ = -112-4·(-6)·10
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{-(-11)-19}{2*-6}=\frac{-8}{-12}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*-6}=\frac{30}{-12}
=-2+1/2
$