(2x+4)(3x-5)+3x(3x+6)=0

Solution for (2x+4)(3x-5)+3x(3x+6)=0 equation:

(2x+4)(3x-5)+3x(3x+6)=0

We multiply parentheses

9x^2+(2x+4)(3x-5)+18x=0

We multiply parentheses ..

9x^2+(+6x^2-10x+12x-20)+18x=0

We get rid of parentheses

9x^2+6x^2-10x+12x+18x-20=0

We add all the numbers together, and all the variables

15x^2+20x-20=0

a = 15; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·15·(-20)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-40}{2*15}=\frac{-60}{30}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+40}{2*15}=\frac{20}{30}
=2/3
$