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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+2x-20=0

a = 6; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·6·(-20)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*6}=\frac{-24}{12}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*6}=\frac{20}{12}
=1+2/3
$